In the given figure,ABCD is a trapezium such that $A L ⊥ D C$ and $B M ⊥ D C$ .If AB=7 cm,BC=AD=5 cm and AL=BM=4 cm then ar(trap.ABCD)=?

(a) $24 c m^{2}$

(b) $40 c m^{2}$

(c) $55 c m^{2}$

(d) $27.5 c m^{2}$

Open in App

Solution

(b) $40 c m^{2}$

In right angled triangle MBC, we have:
MC = √5^2 - 4^2 = √9 = 3 cm
In right angled triangle ADL, we have:
DL = √5^2 - 4^2 = √9 = 3 cm
Now, CD = ML + MC + LD = 7 + 3 + 3 = 13 cm
∴ Area of the trapezium =1/2 × (sum of parallel sides) × distance between them
= 1/2 × ( 13 + 7) × 4
= $40 c m^{2}$

Suggest Corrections

Similar questions

Q. In the given figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

IN the figure, ABCD is a trapezium in Which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm.
Find the value fo x and area of trapezium ABCD.

Q. In the adjoining figure, ABCD is a trapezium in which AB || DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the length of DC and hence, find the area of trap. ABCD.

Q. In the given figure, L and M are the mid- points of AB and BC respectively.

(i) If AB = BC, prove that AL = MC.
(ii) If BL = BM, prove that AB = BC.

Hint
(i)

A B = B C \Rightarrow \frac{1}{2} A B = \frac{1}{2} B C \Rightarrow A L = M C

.
(ii)

B L = B M \Rightarrow 2 B L = 2 B M \Rightarrow A B = B C

A B C D

is a quadrilateral such that

A B = 5 c m, B C = 4 c m; C D = 7 c m, A D = 6 c m

and diagonal

B D = 5 c m

, then the

A (□ A B C D)

4 (3 + \sqrt{6}) c m^{2}

Join BYJU'S Learning Program

In the given figure,ABCD is a trapezium such that AL⊥DC and BM⊥DC.If AB=7 cm,BC=AD=5 cm and AL=BM=4 cm then ar(trap.ABCD)=? (a) 24 cm2 (b) 40 cm2 (c) 55 cm2 (d) 27.5 cm2

In the given figure,ABCD is a trapezium such that $A L ⊥ D C$ and $B M ⊥ D C$ .If AB=7 cm,BC=AD=5 cm and AL=BM=4 cm then ar(trap.ABCD)=?

(a) $24 c m^{2}$

(b) $40 c m^{2}$

(c) $55 c m^{2}$

(d) $27.5 c m^{2}$