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Right option is (d) 22.36 cm

The explanation is: The figure ACCORDING to the given data is:

BD = 10cm and AB = 5cm

Now, in ∆ADB

AD^2 = AB^2 + BD^2

AD^2 = 5^2 + 10^2

AD^2 = 25 + 100

AD^2 = 125

AD = √125 = 11.18 cm

Now, in ∆ABD and ∆CBD

BD = BD(Common SIDE)

∠ADB = ∠CDB(Both 90°)

AB = BC(Given)

∆ABD ≅ ∆CDB (RHS CONGRUENCY)

Therefore, AD = DC

AC = AD + DC = 2AD = 2 × 11.18cm = 22.36cm

Right answer is (a) 6.40 CM

To elaborate: ABCD is a rhombus. The length of DIAGONAL is 10 cm and the length of other diagonal is 8 cm.

Since, diagonals of a rhombus BISECT each other. Therefore, AE = 4CM and DE = 5 cm

Now, in ∆AED

AD^2 = AE^2 + DE^2

AD^2 = 4^2 + 5 ^2(Since, AD is the altitude of the TRIANGLE it will bisect BC)

AD^2 = 16 + 25

AD^2 = 41

AD = √41 cm = 6.40 cm

The correct answer is (a) 4.4 cm

Easiest explanation: We know that the RATIO of areas of similar triangles is equal to the ratio of the squares of their corresponding altitudes.

Here the AREA of ∆ABC is 100cm^2 and area of ∆PQR is 64cm^2. Also, AD = 5.5 cm

According to the theorem,

\(\FRAC {area \, of \, triangle \, \triangle ABC}{area \, of \, triangle \, \triangle PQR}=(\frac {AD}{PS})\)^2

\(\frac {100}{64}=(\frac {5.5}{PS})\)^2

\(\sqrt {\frac {100}{64}}=\frac {5.5}{PS}\)

\(\frac {5.5}{PS}=\frac {10}{8}\)

PS = \(\frac {5.5}{10} × 8 = \frac {44}{10}\) = 4.4

The correct answer is (d) \(\FRAC {40}{3}\)

Easy explanation: We know that the ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides.

Here the area of ∆ABC is 64cm^2 and area of ∆PQR is 81cm^2. Also, QR = 15 cm



According to the theorem,

\(\frac {area \, of \, triangle \, \triangle ABC}{area \, of \, triangle \, \triangle PQR}=(\frac {BC}{QR})\)^2

\(\frac {64}{81}=(\frac {BC}{15})\)^2

\(\sqrt {\frac {64}{81}}=\frac {BC}{15}\)

\(\frac {BC}{15}=\frac {8}{9}\)

BC = \(\frac {8}{9}\) × 15 = \(\frac {4}{3}\)

Right option is (c) 5√2 cm

Best explanation: The diagram according to the given data is:

ABCD is a square INSCRIBED in a circle of radius 5 cm.

Now, JOINING the diagonals of the square we get

The diagonals intersect at E. We know that the diagonals of square are PERPENDICULAR to each other.

In ∆AED, using Pythagoras Theorem,

AD^2 = DE^2 + AE^2

DE and EA are the radius of the circle, ∴ DE = EA

AD^2 = 2DE^2

AD^2 = 2 × 5^2 = 2 × 25 = 50

AD^2 = 50

AD^2 = 125

AD = √50 = 5√2 cm

Correct answer is (b) 225 : 529

The BEST explanation: We know that the RATIO of AREAS of similar triangles is equal to the ratio of the squares of their CORRESPONDING angle bisectors.

Here, AD=1.5 cm and PS=2.3 cm

According to the THEOREM,

\(\frac {area \, of \, \triangle ABC}{area \, of \, \triangle PQR}=(\frac {AD}{PS})\)^2

\(\frac {area \, of \, triangle \, ABC}{area \, of \, triangle \, PQR}=(\frac {1.5}{2.3})\)^2

\(\frac {area \, of \, \triangle ABC}{area \, of \, \triangle PQR}=\frac {2.25}{5.29}=\frac {225}{529}\)

Correct choice is (a) AB = 89, AC = 80, BC = 39

The best I can EXPLAIN: In (a), applying Pythagoras Theorem,

AC^2 + BC^2 = 80^2 + 39^2 = 7921 = AB^2

Hence, this is a right angled TRIANGLE.

Now, in (b) applying Pythagoras Theorem,

AC^2 + BC^2 = 50^2 + 45^2 = 4525 ≠ AB^2

Hence, this is not a right angled triangle.

Now, in (c) applying Pythagoras Theorem,

AB^2 + BC^2 = 20^2 + 21^2 = 841 ≠ AC^2

Hence, this is not a right angled triangle.

Now, in (d) applying Pythagoras Theorem,

AB^2 + BC^2 = 32^2 + 20^2 = 1424 ≠ AC^2

Hence, this is not right angled triangle.

Correct choice is (c) 121 : 81

Best explanation: We know that the ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding altitudes.

Here, AB=5.2cm and AD=4.5 cm

Now, In ∆ACD & ∆ACB

∠ACD = ∠ACB(Both 90°)

AC = AC(Common side)

∠BAC = ∠DAC( AC is the angle bisector)

Hence, ∆ACD ∼ ∆ACB(By, ASA similarity test)

According to the theorem,

\(\FRAC {area \, of \, \triangle ABC}{area \, of \, \triangle ADC}=(\frac {AB}{AD}) \)^2

\(\frac {area \, of \, \triangle ABC}{area \, of \, \triangle ADC}=(\frac {5.5}{4.5}) \)^2

\(\frac {area \, of \, \triangle ABC}{area \, of \, \triangle PQR}=\frac {30.25}{20.25}=\frac {121}{81} \)

The correct choice is (d) 10 cm

Best explanation: ABCD is a rhombus. The SIDE of the rhombus is 13 cm and the length of one of its diagonal is 24 cm.

Let the length of other diagonal be 2x cm.

Since, diagonals of a rhombus bisect each other. Therefore, AE = x cm and DE = 12 cm

Now, in ∆AED

AD^2 = AE^2 + DE^2

13^2 = x^2 + 12^2(Since, AD is the altitude of the triangle it will bisect BC)

x^2 = 169 – 144

x^2 = 25

x = √25 = 5 cm

AC = 2 × AE = 2 × 5 = 10 cm

Right option is (a) True

Easiest EXPLANATION: ∆ABC and ∆PQR are EQUIANGULAR because their CORRESPONDING ANGLES i.e. ∠B, ∠Q are equal.

The correct option is (B) 32.87 m

Easy explanation: Here, AB is the building and AC is the ladder.

Now, the BC can be found out by Pythagoras Theorem.

AC^2 = AB^2 + BC^2

35^2 = 12^2 + BC^2

1225 = 144 + BC^2

BC^2 = 1225 – 144 = 1081

BC = √1081 = 32.87 m

The DISTANCE between the foot of the ladder and the building is 32.87 m.

The correct answer is (d) \(\FRAC {5}{11}\)

For explanation I would SAY: We know that the ratio of areas of SIMILAR triangles is equal to the ratio of the squares of their corresponding medians.

Here the area of ∆ABC is 25cm^2 and area of ∆PQR is 121cm^2. Also, PS = 10 cm

According to the theorem,

\(\frac {area \, of \, triangle \, \triangle ABC}{area \, of \, triangle \, \triangle PQR}=(\frac {AD}{PS})\)^2

\(\frac {25}{121}=(\frac {AD}{10})\)^2

\(\sqrt {\frac {25}{121}}=\frac {AD}{10}\)

\(\frac {AD}{10}=\frac {5}{11}\)

PS = \(\frac {5}{11} × 10 = \frac {50}{11}\)

Right answer is (b) 7.794 cm

The best I can explain: Here, ABC is an EQUILATERAL triangle and AD is the altitude of the triangle.

Now, in ∆ADB

AB^2 = AD^2 + BD^2

9^2 = AD^2 + 4.5^2(Since, AD is the altitude of the triangle it will bisect BC)

AD^2 = 81 – 20.25

AD^2 = 60.75

AD = √60.75 = \(\frac {9\sqrt {3}}{2}\) cm = 7.794 cm

The CORRECT answer is (d) 400 : 9

For explanation I would say: We know that the ratio of areas of SIMILAR triangles is EQUAL to the ratio of the squares of their corresponding altitudes.

Here, AD = 10cm and PS = 5.5 cm

According to the theorem,

\(\frac {area \, of \, \triangle ABC}{area \, of \, \triangle PQR}=(\frac {AD}{PS})\)^2

\(\frac {area \, of \, \triangle ABC}{area \, of \, \triangle PQR}=(\frac {10}{5.5})\)^2

\(\frac {area \, of \, \triangle ABC}{area \, of \, \triangle PQR}=\frac {100}{2.25}=\frac {400}{9}\)

The CORRECT ANSWER is (b) AA test

Explanation: SINCE, the TWO angles of the triangles are equal; therefore the triangles are similar according to the ANGLE – Angle test of similarity.

Correct option is (b) 32 cm

For explanation: We know that the RATIO of the perimeters of similar triangles is the same as the ratio of their corresponding sides.

∴ \(\frac {PERIMETER \, of \, \TRIANGLE ABC}{Perimeter \, of \, \triangle PQR} = \frac {AB}{PQ}\)

\(\frac {64}{24}=\frac {AB}{12}\)

AB = \(\frac {64\times 12}{24}\) = 32 cm

Right choice is (a) AAA TEST

Best EXPLANATION: Since, the angles between the two triangles are equal; THEREFORE the two triangles are SIMILAR ACCORDING to the Angle – Angle – Angle test of similarity.

Right choice is (B) False

Easiest explanation: RATIO of similar triangles is equal to the square of their sides.

Since, ∆ABC ∼ ∆PQR, so they are equiangular and their sides are proportional.

∴ \(\frac {AB}{PQ}=\frac {AC}{PR}=\frac {BC}{QR}\), ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

\( \frac {area \, of \, triange \, ABC}{area \, of \, triange \, PQR}=\frac {\frac {1}{2} \times BC \times AD}{\frac {1}{2} \times QR \times PS}=\frac {BC \times AD}{QR \times PS}\)

Now, ∆ADB and ∆PS,

∠ABD = ∠PQS

∠B = ∠Q

Hence, ∆ABC ∼ ∆PQS

So, \( \frac {AD}{PS}=\frac {AB}{PQ}\)

But, \( \frac {AB}{PQ}=\frac {BC}{QR}\)

∴ \( \frac {AD}{PS}=\frac {BC}{QR}\)

\( \frac {area \, of \, triange \, ABC}{area \, of \, triange \, PQR}=\frac {\frac {1}{2} \times BC \times AD}{\frac {1}{2} \times QR \times PS}=\frac {BC^2}{QR^2}\)

Right choice is (a) 19 : 23

The best I can explain: We KNOW that the ratio of areas of SIMILAR triangles is equal to the ratio of the squares of their corresponding sides.

Here the ratio of areas of two similar triangles is 361:529.



According to the theorem,

\(\frac {area \, of \, triangle \, \triangle ABC}{area \, of \, triangle \, \triangle DEF}=(\frac {AB}{DE})\)^2

\(\frac {361}{529}=(\frac {AB}{DE})\)^2

\(\sqrt {\frac {361}{529}}=\frac {AB}{DE}\)

\(\frac {AB}{DE}=\frac {19}{23}\)