

InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
201. |
In `DeltaABC,"if "angleB=76^(@)andangleC=48^(@)"find "angleA`. |
Answer» Correct Answer - `angleA=56^(@)` | |
202. |
Two circles with different radii are similar. |
Answer» Two rectangles are similar if their corresponding sides are proportional. |
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203. |
Two circles are alwaysA) congruentB) equalC) similarD) can’t say |
Answer» Correct option is C) similar |
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204. |
Which of the following is not the sides of a right triangle ? (A) 3, 4, 5 (B) 5, 12, 13 (C) 7, 24, 25 (D) 4, 5, 6 |
Answer» Correct option is (D) 4, 5, 6 Sides of right triangle will follow Pythagoras theorem which is \((\text{Hypotenuse})^2=(\text{Base})^2+(\text{Altitude})^2\) (A) \(\because\) \(3^2+4^2=9+16\) \(=25=5^2\) \(\therefore\) 3, 4 & 5 are sides of a right triangle. (B) \(\because\) \(5^2+12^2=25+144\) \(=169=13^2\) \(\therefore\) 5, 12 & 13 are sides of a right triangle. (C) \(\because\) \(7^2+24^2=49+576\) \(=625=25^2\) \(\therefore\) 7, 24 & 25 are sides of a right triangle. (D) \(\because\) \(4^2+5^2=16+25\) \(=41\neq36\) \(\therefore\) \(4^2+5^2\neq6^2\) \(\therefore\) Sides 4, 5 & 6 do not form a right triangle. Correct option is: (D) 4, 5, 6 |
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205. |
Find the length of each side of a rhombus whose diagonals are 24 cm and 10 m long. |
Answer» Correct Answer - 13 cm | |
206. |
The lengths of the diagonals of a rhombus are 24 cm and 10 cm. The length of each side of the rhombus isA. 12 cmB. 13 cmC. 14 cmD. 17 cm |
Answer» Correct Answer - B | |
207. |
In Fig, DE || BC, If DE = 4 m, BC = 6 cm and Area (ΔADE) = 16 cm2, find the area of ΔABC. |
Answer» Given, DE ∥ BC. In ΔADE and ΔABC We know that, ∠ADE = ∠B [Corresponding angles] ∠DAE = ∠BAC [Common] Hence, ΔADE ~ ΔABC (AA Similarity) Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we have, \(\frac{Ar(ΔADE)}{Ar(ΔABC)}\) = \(\frac{DE^2}{BC^2}\) \(\frac{16}{Ar(ΔABC)}\) = \(\frac{42}{62}\) ⇒ Ar(ΔABC) = \(\frac{(62 \times 16)}{42}\) ⇒ Ar(ΔABC) = 36 cm2 |
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208. |
State whether the statements are True or FalseIn Fig. 6.29, AD ⊥ BC and AD is the bisector of angle BAC. Then, ΔABD ≅ ΔACD by RHS. |
Answer» In Fig. 6.29, AD ⊥ BC and AD is the bisector of angle BAC. Then, ΔABD ≅ ΔACD by RHS. False |
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209. |
△ABC and △AMP are two right triangles right angled at B and M respectively. Prove that (i) △ABC ~ △AMP (ii) \(\frac{CA}{PA}=\frac{BC}{MP}\) |
Answer» Given: △ABC; ∠B = 90° AAMP; ∠M = 90° R.T.P : i) △ABC ~ △AMP Proof: In △ABC and △AMP ∠B = ∠M [each 90° given] ∠A = ∠A [common] Hence, ∠C = ∠P [∵ Angle sum property of triangles] ∴ △ABC ~ △AMP (by A.A.A. similarity) ii) △ABC ~ △AMP (already proved) \(\frac{AB}{AM}=\frac{BC}{MP}=\frac{CA}{PA}\) [∵ Ratio of corresponding sides of similar triangles are equal] \(\frac{CA}{PA}=\frac{BC}{MP}\) |
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210. |
Find the value of y if the given triangles are similar.(A) 1.5 (B) 2.5 (C) 3.5 (D) 4.5 |
Answer» Correct option is: (B) 2.5 |
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211. |
A man goes 10 m due south and then 24 n due west. How far is from the starting point ? |
Answer» Correct Answer - 26 m | |
212. |
In Fig, DE || BC, If DE: BC = 3: 5. Calculate the ratio of the areas of ΔADE and the trapezium BCED. |
Answer» Given, DE ∥ BC. In ΔADE and ΔABC We know that, ∠ADE = ∠B [Corresponding angles] ∠DAE = ∠BAC [Common] Hence, ΔADE ~ ΔABC (AA Similarity) Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we have, \(\frac{Ar(ΔADE)}{Ar(ΔABC)}\) = \(\frac{DE^2}{BC^2}\) \(\frac{Ar(ΔADE)}{Ar(ΔABC)}\) = \(\frac{3^2}{5^2}\) \(\frac{Ar(ΔADE)}{Ar(ΔABC)}\) = \(\frac{9}{25}\) Assume that the area of ΔADE = 9 x sq units And, area of ΔABC = 25 x sq units So, Area of trapezium BCED = Area of ΔABC – Area of ΔADE = 25x – 9x = 16x Now, \(\frac{Ar(ΔADE)}{Ar(trap\, BCED)}\) = \(\frac{9x}{16x}\) \(\frac{Ar(ΔADE)}{Ar(trap\, BCED)}\) = \(\frac{9}{16}\) |
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213. |
In the given figure, DE || BC and AD : DB = 5 : 4, then = (A) 81/25(B) 5/9(C) 5/4(D) 25/81 |
Answer» Correct option is: (D) \(\frac{25}{81}\) |
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214. |
In a ΔABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm. Find ACA. 4 cm B. 6 cm C. 3 cm D. 8 cm |
Answer» Given AD is the bisector of ∠BAC. AB = 8 cm, DC = 3 cm and BD = 6 cm. We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. ⇒ \(\frac{AB}{AC}\) = \(\frac{BD}{DC}\) ⇒ \(\frac{8}{AC}\) = \(\frac{6}{3}\) ∴ AC = 4 cm |
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215. |
In a ΔABC, AD is the bisector of ∠BAC If AB = 6 cm, AC = 5 cm and BD = 3 cm„ then DC =A. 11.3 cm B. 2.5 cm C. 3 5 cm D. None of these. |
Answer» Given AD is the bisector of ∠BAC. AB = 6 cm, AC = 5 cm and BD = 3 cm. We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. ⇒ \(\frac{AB}{AC}\) = \(\frac{BD}{DC}\) ⇒ \(\frac{6}{5}\) = \(\frac{3}{DC}\) ∴ DC = 2.5 cm |
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216. |
If `Delta ABC~Delta DEF` such that `2AB=DE and BC=6 cm` find EF. |
Answer» Correct Answer - 12 cm | |
217. |
D and E are points on the sides AB and AC respectively of a `Delta ABC` such that `DE||BC`. (i) If `AD=3.6 cm,AB=10 cm and AE=4.5 cm,` find EC and AC. (ii) If `AB=13.3 cm, ac =11. cm, and EC=5.1 cm`, find AD (iii) If `(AD)/(DB)=(4)/(7) and AC=6.6 cm`, find AE. (iv) If `(AD)/(AB)=(8)/(15) and EC=3. cm`, find AE. |
Answer» Correct Answer - `(i) EC =8 cm, AC= 12.5 cm (iii) AD=7.6 cm, (iii) AE=2.4 cm, (iv) AE=4 cm` N/A |
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218. |
In the given figure, `DE||BC` such that `AD=x cm, DB=(3x+4) cm AE= (x+3)cm, and EC =(3x+19) cm`. Find the value of x. |
Answer» Correct Answer - x=2 | |
219. |
In `Delta ABC,DE||BC` so that `AD==2.4 cm, AE= 3.2 cm and EC=4.8 cm`, then , AB= ?A. `3.6 cm`B. 6 cmC. `6.4 cm`D. `7.2 cm` |
Answer» Correct Answer - B | |
220. |
In a ∆ABC, AD is the bisector of ∠A. If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC. |
Answer» If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC. BD/DC = AB/AC BD/3 = 5.6/4 => BD = 4.2 Now, BC = BD + DC = 4.2 + 3 = 7.2 BC is 7.2 cm. |
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221. |
In a ` Delta ABC`, if DE is drawn parallel to BC, cutting AB and AC to D and E respectively such that `AB=7.2 cm, AC=6.4 cm and AD=4.5 cm`. Then, AE=? A. `5.4 cm`B. 4 cmC. `3.6 cm`D. `3.2 cm` |
Answer» Correct Answer - B | |
222. |
In a ∆ABC, AD is the bisector of ∠A. If AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm, find AC. |
Answer» If AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm, find AC. BD/DC = AB/AC Here DC = BC – BD = 6 – 3.2 = 2.8 => DC = 2.8 3.2/2.8 = 5.6/AC => AC = 4.9 cm |
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223. |
The areas of two similar triangles are `81 cm^(2) and 49 cm^(2)` respectively . If the altitude of the triangle is `6.3 cm`, find the corresponding altitude of the other. |
Answer» Correct Answer - `4.9 cm` | |
224. |
When we construct a triangle similar to a given triangle as per given scale factor, we construct on the basis of ………(A) SSS similarity(B) AAA similarity(C) Basic proportionality theorem(D) A and C are correct |
Answer» Correct option is (A) SSS similarity Two triangles are similar if their corresponding sides are proportional and they have two corresponding congruent angles. \(\because\) We have given scale factor (length of sides of triangle) \(\therefore\) We can construct a triangle similar to a given triangle as per given scale factor on the basis of SSS similarity. Correct option is: (A) SSS similarity |
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225. |
In a ∆ABC, AD is the bisector of ∠A. If AB = 10 cm, AC = 14 cm and BC – 6 cm, find BD and DC. |
Answer» If AB = 10 cm, AC = 14 cm and BC – 6 cm, find BD and DC. By angle-bisector theorem BD/DC = AB/AC = 10/14 Let BD = x cm and DC = (6-x) (As BC = 6 cm given) x/(6-x) = 10/14 14x = 10(6 – x) 14x = 60 – 10x 14x + 10x = 60 or x = 2.5 Or BD = 2.5 Then DC = 6 – 2.5 = 3.5 cm |
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226. |
The areas of two similar triangles are 100 cm2 and 49 cm2 respectively. If the altitude of the bigger triangles is 5 cm, find the corresponding altitude of the other. |
Answer» Given: The area of the two similar triangles is 100 cm2 and 49 cm2. And the altitude of the bigger triangle is 5 cm. Required to find: The corresponding altitude of the other triangle We know that, The ratio of the areas of the two similar triangles is equal to the ratio of squares of their corresponding altitudes. ar(bigger triangle)/ ar(smaller triangle) = (altitude of the bigger triangle/ altitude of the smaller triangle)2 (100/ 49) = (5/ altitude of the smaller triangle)2 Taking square root on LHS and RHS, we get (10/ 7) = (5/ altitude of the smaller triangle) = 7/2 Therefore, altitude of the smaller triangle = 3.5 cm |
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227. |
In the fig.given, DE ∥ BC such that AE = (\(\frac{1}{4}\))AC. If AB = 6 cm, find AD. |
Answer» Given: DE∥BC AE = (\(\frac{1}{4}\))AC AB = 6 cm. Required to find: AD. In ΔADE and ΔABC We have, ∠A = ∠A [Common] ∠ADE = ∠ABC [Corresponding angles as AB||QR with PQ as the transversal] ⇒ ΔADE ∼ ΔABC [By AA similarity criteria] Then, \(\frac{AD}{AB}\) = \(\frac{AE}{AC}\) [Corresponding Parts of Similar Triangles are propositional] \(\frac{AD}{6}\)= \(\frac{1}{4}\) 4 x AD = 6 AD =\(\frac{6}{4}\) Therefore, AD = 1.5 cm |
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228. |
In a Δ ABC, D and E are points on the sides AB and AC respectively. Show that DE ∥ BC:If AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm. |
Answer» We have, AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm. So, BD = AB – AD = 5.6 – 1.4 = 4.2 cm And, CE = AC – AE = 7.2 – 1.8 = 5.4 cm It’s seen that, \(\frac{AD}{BD}\) = \(\frac{1.4}{4.2}\) = \(\frac{1}{3}\) \(\frac{AE}{CE}\) = \(\frac{1.8}{5.4}\) = \(\frac{1}{3}\) Thus, \(\frac{AD}{BD}\) = \(\frac{AE}{CE}\) So, by the converse of Thale’s Theorem We have, DE ∥ BC. Hence Proved. |
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229. |
In a ∆ABC, AD is the bisector of ∠A. If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC. |
Answer» If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC. AD bisects ∠A, we can apply angle-bisector theorem in ∆ABC, BD/DC = AB/AC Substituting given values, we get 5.6/DC = 6.4/8 DC = 7 cm |
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230. |
In a Δ ABC, D and E are points on the sides AB and AC respectively. Show that DE ∥ BC:If AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm. |
Answer» We have AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm. So, AD = AB – DB = 10.8 – 4.5 = 6.3 And, CE = AC – AE = 4.8 – 2.8 = 2 It’s seen that, \(\frac{AD}{BD} = \frac{6.3}{4.5} = \frac{2.8}{2.0} = \frac{AE}{CE} = \frac{7}{5}\) So, by the converse of Thale’s Theorem We have, DE ∥ BC. Hence Proved. |
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231. |
D and E are points on the sides AB and AC respectively of a ∆ABC. If AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm. Determine whether DE || BC or not. |
Answer» From figure, D and E are the points on the sides AB and AC of ∆ABC AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm AD/DB = 5.7/9.8 = 3/5 and AE/EC = 4.8/8 = 3/5 => AD/DB = AE/EC => DE || BC |
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232. |
D and E are points on the sides AB and AC respectively of a ∆ABC. If AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm and AC = 10 cm. Determine whether DE || BC or not. |
Answer» From figure, D and E are the points on the sides AB and AC of ∆ABC. AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm and AC = 10 cm. DB = AB – AD = 12 – 7.2 = 4.8 cm and EC = AC – AE = 10 – 6,4 =3.6 cm AD /DB = 7.2/4.8 = 3/2 and AE/EC = 6.4/3.6 = 16/9 => AD/DB ≠ AE/EC => DE is not parallel to BC |
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233. |
ABC is an equilateral triangle of side 2a. Find each of its altitudes. |
Answer» ABC is an equilateral triangle of side 2a. Draw, AD ⊥ BC In ΔADB and ΔADC, we have AB = AC [Given] AD = AD [Given] ∠ADB = ∠ADC [equal to 90°] Therefore, ΔADB ≅ ΔADC by RHS congruence. Hence, BD = DC [by CPCT] In right angled ΔADB, AB2 = AD2 + BD2 (2a)2 = AD2 + a2 ⇒ AD2 = 4a2 - a2 ⇒ AD2 = 3a2 ⇒ AD = √3a |
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234. |
Areas of two similar triangles are 100 cm2 and 64 cm2. If the median of bigger triangle is 10 cm, then the median of the smaller triangle is …………(A) 10 cm (B) 6 cm (C) 4 cm (D) 8 cm |
Answer» Correct option is (D) 8 cm Since, the areas of two similar triangles are in the ratio of the squares of their corresponding medians. \(\therefore\) \(\frac{\text{Area of bigger triangle}}{\text{Area of smaller triangle}}\) \(=(\frac{\text{median of bigger triangle}}{\text{median of smaller triangle}})^2\) \(\therefore\) \((\frac{10}{\text{median of smaller triangle}})^2=\frac{100}{64}\) \(\Rightarrow\) \(\frac{10}{\text{median of smaller triangle}}=\sqrt{\frac{100}{64}}\) \(=\frac{\sqrt{100}}{\sqrt{64}}=\frac{10}{8}\) \(\therefore\) Median of smaller triangle \(=10\times\frac8{10}=8\,cm\) Correct option is: (D) 8 cm |
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235. |
In a Δ ABC, D and E are points on the sides AB and AC respectively. Show that DE ∥ BC:If AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm, and EC = 5.5 cm. |
Answer» We have AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm, and EC = 5.5 cm Now, \(\frac{AD}{BD} = \frac{5.7}{9.5} = \frac{3}{5 }\) And, \(\frac{AE}{CE} = \frac{3.3}{5.3} = \frac{3}{5 }\) Thus, \(\frac{AD}{BD} \) = \(\frac{AE}{CE} \) So, by the converse of Thale’s Theorem We have, DE ∥ BC. Hence Proved. |
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236. |
Sides AB and AC and median AD of a triangle ∆ABC are respectively proportional to sides PQ and PR and mediam PM of another triangle ∆PQR. Show that ∆ABC ~ ∆PQR. |
Answer» Data: Sides AB and AC and median AD of a triangle ∆ABC are respectively proportional to sides PQ and PR and median PM of another triangle ∆PQR. To Prove: ∆ABC ~ ∆PQR In ∆ABC and ∆PQR \(\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}\) Corresponding sides are proportional. Similarity criterion for ∆ is S.S.S ∴ ∆ABC ~ ∆PQR. |
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237. |
Given that ΔABC ~ ΔDEF and their areas are 64 cm2 and. 121 cm2 respectively. If EF = 15.4 cm, then BC =(A) 2.11 cm (B) 21.1 cm (C) 1.21 cm(D) 11.2 cm |
Answer» Correct option is (D) 11.2 cm \(\because\) \(\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=(\frac{BC}{EF})^2\) \((\because\triangle ABC\sim\triangle DEF)\) \(\Rightarrow\) \(\frac{64}{121}=(\frac{BC}{15.4})^2\) \(\Rightarrow\) \(\frac{BC}{15.4}=\sqrt{\frac{64}{121}}=\frac{8}{11}\) \(\Rightarrow BC=\frac8{11}\times15.4\) \(=8\times1.4=11.2\,cm\) Correct option is: (D) 11.2 cm |
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238. |
In the given figure, `angle ABC=90^(@) and BD bot AC`. If `BD=8 cm, AD=4 cm` find CD. |
Answer» Correct Answer - `CD=16 cm` In `Delta DBA and Delta DCB`, we have `angle BDA= angle CDB` and `angle DBA= angle DCB [ "each" =90^(@)-angleA]`. `:. Delta DBA~ Delta DCB`. `:. (BD)/(CD)=(AD)/(BD) rArr CD=(BD)/(AD)` |
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239. |
In given figure, ABC is a triangle right angled at B and BD`bot`AC. If AD=4 cm and CD= 5 cm, then find BD and AB. |
Answer» Given`DeltaABC` in which`angleB=90^(@) and BDbotAC` Also, AD=4 cm and CD= 5 cm In `DeltaADB and DeltaCDB, angleADB=angleCDB` [each equal to `90^(@)`] and `angleBAD=angleDBC`[each equal to `90^(@)-angleC`] `therefore DeltaDBA~DeltaDCB` [by AAA similarity criterion] Then, `(DB)/(DA)=(DC)/(DB)` `rArr DB^(2)=DAxxDC` `rArrDB^(2)=4xx5` `rArrDB=2sqrt5cm` In right angled `DeltaBDC, BC^(2)=BD^(2)+CD^(2)` [by phythogoras therom] `rArrBC^(2)=(2sqrt5)^(2)+(5)^(2)` =20+25=45 `rArrBC=sqrt(45)=3sqrt5` Again, `DeltaDBA~DeltaDCB` `therefore(DB)/(DC)=(BA)/(BC)` `rArr(2sqrt5)/5=(BA)/(3sqrt5)` `therefore BA=(2sqrt5xx3sqrt5)/5=6 cm` Hence, BD=`2sqrt5` cm and AB= 6 cm |
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240. |
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD. |
Answer» Data: D is a point on the side BC of a triangle ∆ABC such that ∠ADC = ∠BAC. To Prove: CA2 = CB × CD Let ∠ADC = ∠BAC = 100° In ∆ABC, If ∠B = 50°,then ∠C = 30° In ∆ADC, If ∠C = 30°. then ∠DAC = 50° In ∆BCP, ∠A= 100°, ∠B= 50°, ∠C= 30° In ∆ADC, ∠ADC = 100. ∠DAC = 50°. ∠ACD = 30° Similarity criterion of ∆ is A.A.A. ∴ In ∆ABC and ∆ADC, \(\frac{CA}{BC} = \frac{DC}{CA}\) ∴ CA × CA = BC × DC ∴ CA2 = BC × DC. |
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241. |
ΔABC ~ ΔDEF and their areas are 16 cm2 and 81 cm2 respectively.If BC = 4 cm, then EF =(A) 5 cm (B) 4 cm (C) 11 cm (D) 9 cm |
Answer» Correct option is (D) 9 cm \(\because\) \(\triangle ABC \sim \triangle DEF\) \(\therefore\) \(\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=(\frac{BC}{EF})^2\) \(\Rightarrow\)\((\frac{BC}{EF})^2=\frac{16}{81}\) (Given \(ar(\triangle ABC)=16\,cm^2\;\&\;ar(\triangle DEF)=81\,cm^2)\) \(\Rightarrow\) \(\frac{BC}{EF}=\sqrt{\frac{16}{81}}=\frac49\) \(\Rightarrow EF=\frac94BC\) \(=\frac94\times4=9\) \((\because BC=4\,cm)\) \(\therefore\) EF = 9 cm Correct option is: (D) 9 cm |
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242. |
In the figure below, PQ || BC. If AP = 3, PB = 4 and AQ = 4, then QC =(A) 7/3(B) 4/3(C) 16/3(D) 3/4 |
Answer» Correct option is (C) 16/3 \(\because\) PQ || BC in \(\triangle ABC\) \(\therefore\) \(\triangle APQ\sim\triangle ABC\) \(\therefore\) \(\frac{AP}{AB}=\frac{AQ}{AC}\) \(\Rightarrow\) \(\frac{AP}{AB-AP}=\frac{AQ}{AC-AQ}\) \(\Rightarrow\) \(\frac{AP}{PB}=\frac{AQ}{QC}\) \(\Rightarrow\) \(QC=AQ.\frac{PB}{AP}\) \(=4.\frac43=\frac{16}3\) \((\because\) AQ = 4, PB = 4 & AP = 3 (given)) Correct option is: (C) \(\frac{16}{3}\) |
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243. |
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. |
Answer» Let `a` is the side of given triangle, then its diagonal will be, `sqrt(a^2+a^2) = sqrt2a` As, all equilateral triangles are similar, ratio of their ares will be squares of ratio of their sides. So, ratio of area of given triangles, `=a^2/(sqrt2a)^2 = a^2/(2a^2) = 1/2` Thus, the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. |
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244. |
In an equilateral triangle ABC (Fig), AD is an altitude. Then 4AD2 is equal to(a) 2BD2 (b) BC2 (c) 3AB2 (d) 2DC2 |
Answer» Correct answer is (c) 3AB2. |
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245. |
In Fig., D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that:(i) \(b^2 = p^2+a+\frac{{a}^2}{4}\)(ii) \(c^2 = p^2 -{ax}+\frac{a^2}{4}\)(iii)\(v^2+c^2 =2p^2+\frac{{a}^2}{2}\) |
Answer» We have D is the midpoint of BC (i) In ΔAEC AC2 = AE2 + EC2 b2 = AE2 + (ED + DC)2 b2 = AD2 + DC2 + 2 x ED x DC (Given BC = 2CD) b2 = p2 + (a/2)2 + 2(a/2)x b2 = p2 + a2/4 + ax b2 = p2 + ax + a2/4 ………….. (i) (ii) In ΔAEB AB2 = AE2 + BE2 c2 = AD2 - ED2 + (BD - ED)2 c2 = p2 - ED2 + BD2+ ED2 - 2BD x ED c2 = P2 + (a/2)2 - 2(a/2)2x c2 = p2 - ax + a2/4 ……………….(ii) (iii) Adding equ. (i)and(ii) we get b2 + c2 = 2p2 + a2/2 |
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246. |
Choose the correct one. In ΔPQR, if PQ = QR and ∠Q = 100°, then ∠R is equal to (a) 40° (b) 80° (c) 120° (d) 50° |
Answer» Correct answer is (a) 40° |
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247. |
In △ PQR, if ∠P – ∠Q = 42° and ∠Q – ∠R = 21°, find ∠P, ∠Q and ∠R. |
Answer» It is given that ∠P – ∠Q = 42o It can be written as ∠P = 42o + ∠Q We know that the sum of all the angles in a triangle is 180o. So we can write it as ∠P + ∠Q + ∠R = 180o By substituting ∠P = 42o + ∠Q in the above equation 42o + ∠Q +∠Q + ∠R = 180o On further calculation 42o + 2 ∠Q + ∠R = 180o 2 ∠Q + ∠R = 180o – 42o By subtraction we get 2 ∠Q + ∠R = 138o …. (i) It is given that ∠Q – ∠R = 21o It can be written as ∠R = ∠Q – 21o By substituting the value of ∠R in equation (i) 2 ∠Q + ∠Q – 21o = 138o On further calculation 3 ∠Q – 21o = 138o 3 ∠Q = 138o + 21o By addition 3 ∠Q = 159o By division ∠Q = 159/3 ∠Q = 53o By substituting ∠Q = 53o in ∠P = 42o + ∠Q So we get ∠P = 42o + 53o By addition ∠P = 95o By substituting ∠Q in ∠Q – ∠R = 21o 53o – ∠R = 21o On further calculation ∠R = 53o – 21o By subtraction ∠R = 32o Therefore, ∠P = 95o, ∠Q = 53o and ∠R= 32o |
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248. |
In ΔPQR ∠P= 2 ∠Q and 2 ∠R =3 ∠Q , calculate the angles of ΔPQR. |
Answer» In ΔPQR ∠P + ∠Q + ∠R = 180° also ∠P:∠Q:∠R = 2∠Q:∠Q: 3/2 ∠Q = 4 : 2 : 3 Sum of the term of the ratio = 4 + 2 + 3 = 9 ∠P = 4/9 x 180° = 80° ∠Q = 2/9 x 180° = 40° ∠R = 3/9 x 180° = 60° |
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249. |
In ΔPQR, ∠R = 90° then hypotenuse is ………………….. (A) PQ (B) PR (C) RP (D) PP |
Answer» Correct option is (A) PQ In \( \triangle PQR, \angle R\) = 90° Then side opposite to \(\angle R\) is hypotenuse of right \( \triangle PQR.\) \(\because\) PQ is side opposite to \(\angle R\) in \( \triangle PQR.\) \(\therefore\) PQ is hypotenuse. Correct option is: (A) PQ |
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250. |
Which one of the following is not a criterion for congruence of two triangles? (a) ASA (b) SSA (c) SAS (d) SSS |
Answer» Correct answer is (b) SSA. |
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