InterviewSolution
Saved Bookmarks
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.
| 1. |
The area bounded by the graph of `y=f(x), f(x) gt0` on [0,a] and x-axis is `(a^(2))/(2)+(a)/(2) sin a +(pi)/(2) cos a ` then find the value of `f((pi)/(2))`. |
|
Answer» Correct Answer - `(1)/(3)` According to the equestion `overset(a)underset(0)intf(x)dx=(a^(2))/(2)+(a)/(2)sin a +(pi)/(2) cos a ` `therefore" "f(a)=(d)/(da)((a^(2))/(2)+(a)/(2)sin a +(pi)/(2)cos a)` `a+(a)/(2) cos a +(1)/(2)sin a -(pi)/(2) sin a` `therefore" "f((pi)/(2))=(1)/(2)` |
|
| 2. |
If the area bounded by the x-axis, the curve `y=f(x), (f(x)gt0)" and the lines "x=1, x=b " is equal to "sqrt(b^(2)+1)-sqrt(2)" for all "bgt1,` then find f(x). |
|
Answer» Correct Answer - `(x)/(sqrt(x^(2)+1))` `"Area "=int_(1)^(b)f(x)dx=sqrt(b^(2+1))-sqrt(2)` Differentiating both sides w.r.t. b, we get `f(b)=(b)/(sqrt(b^(2)+1))` `therefore" "f(x)=(x)/(sqrt(x^(2)+1))` |
|
| 3. |
Find the continuous function `f`where `(x^4-4x^2)lt=f(x)lt=(2x^2-x^3)`such that the area bounded by `y=f(x),y=x^4-4x^2dot`then y-axis, and the line `x=t ,`where `(0lt=tlt=2)`is `k`times the area bounded by `y=f(x),y=2x^2-x^3,y-a xi s ,`and line `x=t(w h e r e0lt=tlt=2)dot` |
|
Answer» Correct Answer - `(1)/(k+1)(x^(4)-kx^(3)+2(k-2)x^(2))` According to the given conditions `overset(t)underset(0)int[f(x)-(x^(4)-4x^(2))]dx=koverset(t)underset(0)int[(2x^(2)-x^(3))-f(x)]dx` Differentiating both sides w.r.t. we get `f(t)-(t^(4)-4t^(2))=k(2t^(2)-t^(3)-f(t))` `"or "(1+k)f(t)=k2t^(2)-kt^(2)+t^(4)-4t^(2)` `rArr" "f(t)=(1)/(k+1)[t^(4)-kt^(3)+(2k-4)t^(2)]` Hence, required f is given by `f(x)=(1)/(k+1)(x^(4)-kx^(3)+2(k-2)x^(2)).` |
|
| 4. |
In any ΔABC, D is a point on BC such that ar (ΔABD) = ar (ΔADC then \(\overline{AD}\) represents. A) Altitude B) Median C) Angle bisector D) Perpendicular Bisector |
|
Answer» Correct option is B) Median |
|
| 5. |
If 1 cm represents 2.5 m then area of a square with 10 m is A) 10 cm2B) 4 cm2 C) 2.5 cm2 D) 1 cm2 |
|
Answer» Correct option is B) 4 cm2 |
|
| 6. |
If 1 cm represents 5 in, what would be an area of 6 cm2 represents ? |
|
Answer» 1 cm2 = 5 m 1 cm2 = 1 cm × 1 cm = 5m × 5m = 25m2 ∴ 6 cm2 = 6 × 25 m2 = 150 m2 |
|
| 7. |
Rajni says 1 sq. m = 1002 sq. cm. Do you agree ? |
|
Answer» No 1 sq. m = 100 cm2 |
|
| 8. |
In the adjacent figure the ratio of ar (ΔAPB) and ar (ΔDQC) is ……………….A) 1 : 2 B) 2 : 1 C) 1 : 1 D) 1 : 3 |
|
Answer» Correct option is C) 1 : 1 |
|
| 9. |
The area of parallelogram ABCD is 36cm2 . Calculate the height of parallelogram ABEF if AB = 4.2 cm. |
|
Answer» Area of □ABCD = 36 cm2 AB = 4.2 cm then □ABCD = AB X Height [ ∵ base x height] 36 = 4.2 x h ∴ h = 36/4.2 But □ ABCD and □ ABEF are on the same base and between the same parallels. ∴ □ABCD = □ABEF □ABEF = base x height = AB x height ∴ height = 36/4.2 = 8.571cm |
|
| 10. |
P is a point in the interior of a parallelogram ABCD. Show that ar (ΔAPB) + ar (ΔPCD) = 1/2 ar(ABCD) (Hint : Through P, draw a line parallel to AB) |
|
Answer» Solution: □ABCD is a parallelogram. P is any interior point. Draw a line \(\overline {XY}\) parallel to AB through P. Now ΔAPB = 1/2 □AXYB ……………(1) [∵ ΔAPB, □AXYB lie on the same base AB and between AB//XY] Also ΔPCD = 1/2 □CDXY ………………… (2) [ ∵ ΔPCD; □CDXY lie on the same base CD and between CD//XY] Adding (1) & (2), we get Δ APB + ΔPCD = 1/2 □AXYB + 1/2 □CDXY = 1/2 [□ AXYB + □ CDXY] [from the fig.) = 1/2 □ABCD Hence Proved. |
|
| 11. |
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (ΔAPB) = ar (ΔBQC). |
|
Answer» ΔAPB and □ABCD are on the same base AB and between the same parallel lines AB//CD. ∴ ΔAPB = 1/2 □ABCD …………… (1) Also ΔBCQ and □BCDA are on the same base BC and between the same parallel lines BC//AD. ∴ ΔBCQ = 1/2 □BCDA …………….. (2) But □ABCD and □BCDA represent same parallelogram. ∴ΔAPB = ΔBCQ [from (1) & (2)] |
|
| 12. |
ABCD is a parallelogram. AE is perpendicular on DC and CF is perpendicular on AD. If AB = 10 cm; AE = 8 cm and CF = 12 cm. Find AD. |
|
Answer» Area of parallelogram = base x height AB x AE = AD x CF ⇒ 10 x 8 = 12 x AD ⇒ AD = 10x8/12 = 6.666 ………. ∴ AD ≅ 6.7 cm |
|
| 13. |
PQRS and ABRS are parallelograms and X is any point on the side BR. Show that ar (PQRS) = ar (ABRS) |
|
Answer» □PQRS and □ABRS are on the same base SR and between the same parallels SR//PB. ∴ □PQRS = □ABRS |
|
| 14. |
In a triangle ABC, E is the midpoint of median AD. Show that ar ΔABE = ar ΔACE |
|
Answer» In ΔABC; AD is a median. ∴ ΔABD = ΔACD …………….. (1) (∵ Median divides a triangle in two equal triangles) Also in ΔABD; BE is a median. ∴ ΔABE = ΔBED = 1/2 ΔABD …………..(2) Also in ΔACD; CE is a median. ∴ ΔACE = ΔCDE = 1/2 ΔACD …………….(3) From (1), (2) and (3); ΔABE = ΔACE (OR) ΔABD = ΔACD [∵ AD is median in ΔABC] 1/2 ΔABD = 1/2 ΔACD [Dividing both sides by 2] ΔABE = ΔAEC [∵ BE is median of ΔABD, CE is median of ΔACD] Hence proved. |
|
| 15. |
In the figure ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that ar (AEDF) = ar (ABCDE) |
|
Answer» □AEDF = □AEDC + ΔACF = □AEDC + ΔABC [ ∵ ΔACF = ΔACB] = area (ABCDE) Hence proved. |
|
| 16. |
The value of the parameter a such that the area bounded by `y=a^(2)x^(2)+ax+1,` coordinate axes, and the line x=1 attains its least value is equal toA. `(1)/(4)` sq. unitsB. `-(1)/(2)` sq. unitsC. `(3)/(4)` sq. unitsD. `-1` sq. units |
|
Answer» Correct Answer - C `a^(2)x^(2)+ax+1` is clearly positive for all real values of x, Area under consideration `A=overset(1)underset(0)int(a^(2)x^(2)+ax +1)dx` `=(a^(2))/(3)+(a)/(2)+1` `=(1)/(6)(2a^(2)+3a+6)` `=(1)/(6)(2(a^(2)+(3)/(2)a+(9)/(16))+6-(18)/(16))` `=(1)/(6)(2(a+(3)/(4))^(2)+(39)/(8))`, which is clearly minimum for `a=-(3)/(4).` |
|
| 17. |
The area bounded by the curve `y=x(1-log_(e)x)` and x-axis isA. `(e^(2))/(4)`B. `(e^(2))/(2)`C. `(e^(2)-e)/(2)`D. `(e^(2)-e)/(4)` |
|
Answer» Correct Answer - A `y=x(1-log_(e)x)` It meets x-axis, if `x(1-loge_(e)x)=0` `therefore" "x=0 or x=e` `therefore" Required area "=overset(e)underset(0)intx(1-log_(e)x)dx` `=overset(e)underset(0)intxdx -overset(e)underset(0)intx log x dx` `=[(x^(2))/(2)]_(0)^(e)-[(x^(2))/(2)log_(e)x]_(0)^(e)+overset(e)underset(0)int(x)/(2)dx` `=(e^(2))/(2)-[(e^(2))/(2)-underset(xrarr0)lim(x^(2))/(2)log_(e)x]+(e^(2))/(4)` `=(e^(2))/(4)` |
|
| 18. |
In ΔABC, ∠ABC = 90°; AD = DC; AB =12 cm, BC = 6.5 cm. Find the area of ΔADB |
|
Answer» ΔADB = 1/2 ΔABC [ ∵ AD is a median of ΔABC] 1/2 = [1/2 AB x BC] = 1/4 x 12 x 6.5 = 19.5 cm2 |
|
| 19. |
Find the area of the circle if its circumference is 88 cm. |
|
Answer» Circumference of the circle = 88 cm … [Given] Circumference of the circle = 2πr ∴ 88 = 2 x (22/7) x r ∴ r = (88 x 7/2 x 22) ∴ r = 14cm Area of the circle = πr2 = (22/7) x (14)2 = (22/7) x 14 x 14 = 22 x 2 x 14 = 616 sq. cm ∴ The area of circle is 616 Sq cm |
|
| 20. |
Draw a big enough parallelogram ABCD on a paper as shown in the figure. Draw perpendicular AE on side BC. Cut the right angled ∆AEB. Join it with the remaining part of ₹ABCD as shown in the figure. The new figure formed is a rectangle. The rectangle is formed from the parallelogram. So, areas of both the figures are equal. Base of parallelogram is one side (length) of the rectangle and its height is the other side (breadth) of the rectangle. ∴ Area of a parallelogram = base × height |
|
Answer» Draw a big enough parallelogram ABCD on a paper as shown in the figure. Draw perpendicular AE on side BC. Cut the right angled ∆AEB. Join it with the remaining part of ₹ABCD as shown in the figure. The new figure formed is a rectangle. The rectangle is formed from the parallelogram. So, areas of both the figures are equal. Base of parallelogram is one side (length) of the rectangle and its height is the other side (breadth) of the rectangle. ∴ Area of a parallelogram = Area of a rectangle = length × breadth = base × height |
|
| 21. |
Some measures are given in the figure, find the area of ABCD. |
|
Answer» A(ABCD) = A(∆BAD) + A(∆BDC) In ∆BAD, m∠BAD = 90°, l(AB) = 40m, l(AD) = 9m A(∆BAD) = (1/2) x product of sides forming the right angle = (1/2) x l(AB) x l(AD) = (1/2) x 40 x 9 = 180 sq. m In ∆BDC, l(BT) = 13m, l(CD) = 60m A(∆BDC) = (1/2) x base x height = (1/2) x l(CD) x l(BT) = (1/2) x 60 x 13 = 390 sq. m A (ABCD) = A(∆BAD) + A(∆BDC) = 180 + 390 = 570 sq. m ∴ The area of ABCD is 570 sq.m. |
|
| 22. |
Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area. |
|
Answer» Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm and its height is 4.2 cm. Area of a trapezium = \(\cfrac{1}{2}\)x sum of lengths of parallel sides x height = \(\cfrac{1}{2}\)x (8.5 + 11.5) x 4.2 = \(\cfrac{1}{2}\) x 20 x 4.2 = 10 x 4.2 = 42 sq. cm ∴ The area of the trapezium is 42 sq. cm. |
|
| 23. |
Area of a parallelogram is 83.2 sq.cm. If its height is 6.4 cm, find the length of its base. |
|
Answer» Given, area of a parallelogram = 83.2 sq.cm, height = 6.4 cm Area of a parallelogram = base × height ∴ 83.2 = base × 6.4 ∴ base = 83.2/6.4 = 13 cm ∴ The length of the base of the parallelogram is 13 cm. |
|
| 24. |
Radii of the circles are given below, find their areas. i. 28 cm ii. 10.5 cm iii. 17.5 cm |
|
Answer» i. Radius of the circle (r) = 28 cm … [Given] Area of the circle = πr2 = (22/7) x (28)2 = (22/7) x 28 x 28 = 22 x 4 x 28 = 2464 sq. cm ii. Radius of the circle (r) = 10.5 cm … [Given] Area of the circle = πr2 = (22/7) x (10.5)2 = (22/7) x 10.5 x 10.5 = 22 x 1.5 x 10.5 = 346.5 sq. cm iii. Radius of the circle (r) = 17.5 cm … [Given] Area of the circle = πr2 = (22/7) x (17.5)2 = (22/7) x 17.5 x 17.5 = 22 x 2.5 x 17.5 = 962.5 sq. cm |
|
| 25. |
Area of a parallelogram is the product of its A) consecutive sides B) opposite sides C) base and height D) diagonals |
|
Answer» C) base and height |
|
| 26. |
Area of a parallelogram is 64 cm2 with base 16 cm. Then its height is A) 8 cm B) 4 cm C) 6 cmD) 5 cm |
|
Answer» Correct option is B) 4 cm |
|
| 27. |
A polygon can be divided into …………………… regions. A) Square B) Rectangular C) Circular D) Triangular |
|
Answer» D) Triangular |
|
| 28. |
If □PQRS = 302 cm then PM =A) 6 cm B) 3 cm C) 20 cm D) None |
|
Answer» Correct option is B) 3 cm |
|
| 29. |
In ΔPQR, A, B and C are the mid points of the sides, then □AQRC =A) 1/2 ΔPQR B) 1/4 ΔPQRC) 3/4 ΔPQR D) 2/3 ΔPQR |
|
Answer» Correct option is C) 3/4 ΔPQR |
|
| 30. |
In the figure, area of □ABCD if the area of ΔADE is 24 cm2 A) 24 cm2C) 48 cm2 B) 36 cm2 D) 12 cm2 |
|
Answer» Correct option is C) 48 cm2 |
|
| 31. |
In the figure, AB // CF then □ABCD =A) □ABED B) □ABEF C) □ABCF D) □ABCE |
|
Answer» Correct option is B) □ABEF |
|
| 32. |
In the given figure, ΔABC, D, E, F are the mid points of sides BC, CA and AB respectively. Then BDEF is a A) square B) rectangle C) rhombus D) parallelogram |
|
Answer» D) parallelogram |
|
| 33. |
In the above figure, ar ΔPTS = A) ΔPQT B) ΔQRT C) 2ΔTSR D) 1/2 ΔPQT |
|
Answer» Correct option is B) ΔQRT |
|
| 34. |
□ABCD is a parallelogram, then ΔAEH + ΔBEF + ΔCGF + ΔDGB is (where E, F, G and H are the mid points of the sidesA) □ABCD B) □EFGH C) 1/2 □EFGH D) 2 □EFGH |
|
Answer» Correct option is B) □EFGH |
|
| 35. |
In the figure, ar ΔPQR = 15 cm2 then ar ΔPQS =A) 7.5 cm2 B) 15 cm2 C) 30 cm2 D) 10 cm2 |
|
Answer» Correct option is B) 15 cm2 |
|
| 36. |
The area enclosed by the curve `C:y=xsqrt(9-x^(2))(xge0)` and the x-axis is___. |
|
Answer» Correct Answer - 9 Required area `A=overset(3)underset(0)intxsqrt(9-x^(2))dx," Put "9-x^(2)=t^(2)rArr-2x" "dx =2t" "dt` `therefore" "A=overset(3)underset(0)intt^(2)dt=9` |
|
| 37. |
Consider two functions `f (x) ={[x] , -2 leq x leq -1 and |x|+1 , -1 lt x leq 2 and g(x)={[x], -pi leq x lt 0 and sin x and 0 leq x leq pi`, where [.] denotes the greatest integer function.A. `(sqrt(3))/(4)+(pi)/(6)` sq. unitsB. `(sqrt(3))/(2)+(pi)/(6)` sq. unitsC. `(sqrt(3))/(4)-(pi)/(6)` sq. unitsD. `(sqrt(3))/(2)-(pi)/(6)` sq. units |
|
Answer» Correct Answer - A `A=2overset(1//2)underset(0)intsqrt(1-x^(2))dx` `=2[(x)/(2)sqrt(1-x^(2)):|_(0)^((1)/(2))+(1)/(2) sin^(-1)x:|_(0)^((1)/(2))]` `= (sqrt(3))/(4)+(p)/(6)` sq. unit. |
|
| 38. |
Let `f:[-1,2]vec[0,oo)`be a continuous function such that `f(x)=f(1-x)fora l lx in [-1,2]dot`Let `R_1=int_(-1)^2xf(x)dx ,`and `R_2`be the area of the region bounded by `y=f(x),x=-1,x=2,`and the `x-a xi s`. Then`R_1=2R_2`(b) `R_1=3R_2``2R_1`(d) `3R_1=R_2`A. `R_(1)=2R_(2)`B. `R_(1)=3R_(2)`C. `2R_(1)=R_(2)`D. `3R_(1)=R_(2)` |
|
Answer» Correct Answer - C `R_(1)=int_(-1)^(2)xf(x)dx " "...(i)` Using ` int_(a)^(b)f(x)dx =int_(a)^(b)f(a+b-x)dx ` ` R_(1)=int_(-1)^(2)(1-x)f(1-x)dx ` ` therefore R_(1)=int_(-1)^(2)(1-x)f(x)dx " "...(ii) ` `[f(x)=f(1-x), " given "]` Given, ` R_(2) ` is area bounded by ` f(x),x=-1 " and " x=2 .` ` therefore R_(2)=int_(-1)^(2)f(x)dx " "...(iii)` On adding Eqs.(i) and (ii), we get `2R_(1)=int_(-1)^(2)f(x)dx " "...(iv)` From Eqs.(iii) and (iv), we get ` 2R_(1)=R_(2) ` |
|
| 39. |
If the line `x= alpha ` divides the area of region `R={(x,y)in R^(2): x^(3) le y le x ,0 le x le 1 } ` into two equal parts, thenA. ` 2 alpha^(4)-4alpha^(2)+1=0`B. `alpha^(4)+4alpha^(2)-1=0`C. `(1)/(2) lt alpha lt 1 `D. `0 lt alpha le (1)/(2)` |
|
Answer» Correct Answer - A::C ` int_(0)^(1)(x-x^(3))dx= 2int_(0)^(alpha)(x-x^(3))dx ` `(1)/(4)=2((alpha^(2))/(2)-(alpha^(4))/(4))` `2 alpha^(4)-4 alpha^(2)+1=0 ` ` rArr alpha^(2)=(4-sqrt(16-8))/(4) " "( because alpha in (0,1))` ` alpha^(2)=1-(1)/(sqrt(2))` |
|
| 40. |
Let `C_(1) and C_(2)` be the graphs of the functions `y=x^(2) and y=2x,` respectively, where `0le x le 1." Let "C_(3)` be the graph of a function y=f(x), where `0lexle1, f(0)=0.` For a point P on `C_(1),` let the lines through P, parallel to the axes, meet `C_(2) and C_(3)` at Q and R, respectively (see figure). If for every position of `P(on C_(1)),` the areas of the shaded regions OPQ and ORP are equal, determine the function f(x). |
|
Answer» Let P be on `C_(1),y=x^(2) be (t,t^(2))` `therefore" y co-ordinate of Q is also "t^(2)` `"Now, Q on y =2x where "y=t^(2)` `therefore" "x=t^(2)//2` `therefore" "Q((t^(2))/(2),t^(2))` For point R, x=t and it is on y=f(x) `therefore" "R(t,f(t))` Given that, Area OPQ = Area OPR `rArr" "int_(0)^(t^(2))(sqrt(y)-(y)/(2))dy=int_(0)^(t)(x^(2)-f(x))dx` Differentiating both sides w.r.t. t, we get `(sqrt(t^(2))-(t^(2))/(2))(2t)=t^(2)-f(t)` `rArr" "f(t)=t^(3)-t^(2)` `rArr" "f(x)=x^(3)-x^(2)` |
|
| 41. |
If the straight line x=b divide the area enclosed by `y=(1-x)^(2),y=0 " and " x=0 " into two parts " R_(1)(0le x le b) " and " R_(2)(b le x le 1) " such that " R_(1)-R_(2)=(1)/(4). ` Then, b equalsA. `(3)/(4)`B. `(1)/(2)`C. `(1)/(3)`D. `(1)/(4)` |
|
Answer» Correct Answer - B Here, area between 0 to b is `R_(1)` and b to 1 is ` R_(2).` `therefore int_(0)^(b)(1-x)^(2)dx-int_(b)^(1)(1-x)^(2)dx = (1)/(4) ` `rArr [((1-x)^(3))/(-3)]_(0)^(b)-[((1-x)^(3))/(-3)]_(b)^(1)=(1)/(4) ` `rArr -(1)/(3)[(1-b)^(3)-1]+(1)/(3)[0-(1-b)^(3)]=(1)/(4) ` ` rArr -(2)/(3)(1-b)^(3)=-(1)/(3)+(1)/(4)=-(1)/(12) rArr (1-b)^(3)=(1)/(8) ` ` rArr (1-b)=(1)/(2) rArr b=(1)/(2) ` |
|
| 42. |
The field of the kalyan is in the form of quadrilateral. The diagonal of this field is 220 m and the perpendiculars dropped on it from the remaining opposite vertices are 80 m and 130 m respectively. Find the area of the field. |
|
Answer» Area of quadrilateral shaped farm = 1/2 x diagonal x (Sum of perpendiculars drawn from opposite vertices) = 1/2 x 220 x (80 + 130) = 1/2 x 220 x 210 = 23100 m2 |
|
| 43. |
Area of a parallelogram is A) base × altitude B)1/2 × base × height C) Product of the diagonals D) Half of the product of the diagonals |
|
Answer» A) base × altitude |
|
| 44. |
Area of a Rhombus is A) 1/2 × base x height B) Product of the diagonals C) Half of the product of the diagonalsD) 1/2 × sum of the parallel sides × distance between them |
|
Answer» C) Half of the product of the diagonals |
|
| 45. |
Find the area of the combined rhombus shaped tiles as given in the figure. |
|
Answer» Area of rhombus shaped tiles = 2[1/2 x (4.5×9)] Area of rhombus = 1/2 x Product of diagonals = 40.5 cm2 |
|
| 46. |
Length between the opposite vertex of the rhombus shaped plot are 12.5 m and 10.4 m. Find the total cost of making this plot as a flat surface if the cost of making a flat surface per square meter is Rs. 180. |
|
Answer» Area of rhombus shaped plot = 1/2 x Product of diagonals = 1/2 x 12.5 x 10.4 = 65 meter2 Cost of making plane this plot = 65 x 180 = Rs 11,700 |
|
| 47. |
Area of a rhombus with diagonals 6 cm and 10 cm is A) 60 cm2 B) 30 cm2 C) 16 cm2 D) 15 cm2 |
|
Answer» Correct option is B) 30 cm2 |
|
| 48. |
In the figure line AB is parallel to CD. If AB = 5cm, AC = 4cm and ∠CAB = 90°.a. Calculate the area of triangle ABC.b. How much is the area of triangle ∆ ABD?Write reason |
|
Answer» a. Area of ∆ ABC = \(\frac{1}{2}\) x 5 x 4 = 10 sq.cm b. Area of ∆ ABD = Area of ∆ ABC Triangles with the same base and the third vertex on a line parallel to the base, have the same area. Area of ∆ ABD = 10 sq.cm |
|
| 49. |
Lengths of the diagonals of a rhombus are 15 cm and 24 cm, find its area. |
|
Answer» Lengths of the diagonals of a rhombus are 15 cm and 24 cm. Area of a rhombus = \(\cfrac{1}{2}\)× product of lengths of diagonals = \(\cfrac{1}{2}\) × 15 × 24 = 15 × 12 = 180 sq.cm ∴ The area of the rhombus is 180 sq. cm. |
|
| 50. |
Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area. |
|
Answer» Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm. Area of a rhombus = (1/2) × product of lengths of diagonals = (1/2) × 16.5 × 14.2 = 16.5 × 7.1 = 117.15 sq cm ∴ The area of the rhombus is 117.15 sq. cm. |
|