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101.

In the figure, diagonals AC and BD of a trapezium ABCD with AB//DC intersect each other at ‘O’. Prove that ar (ΔAOD) = ar (ΔBOC)

Answer»

Given that AB // CD 

Now ΔADC and ΔBCD are on the same base and between the same parallels AB // CD. 

∴ ΔADC = ΔBCD 

⇒ ΔADC – ΔCOD = ΔBCD – ΔCOD 

⇒ ΔAOD = ΔBOC [from the figure]
Discussion
102.

Area of a rhombus ABCD is 264 cm2. If length of its one diagonal AC = 24 cm then find length of diagonal BD.

Answer»

We know that,

Area of rhombus = 1/2 x product of diagonals

Given: Area of rhombus = 264 cm2, diagonal AC = 24 cm

To find = diagonal BD ,

∴ 264 = 1/2 x 24 x BD

BD = \(\frac{264 \times 2}{24}\) = 22 cm.
Discussion
103.

ABCD is a parallelogram. The diagonals AC and BD intersect at ‘O’. If ar (ΔAOB) = 3.5 cm2 , then ar (ΔABC) =A) 10.5cm2B) 21 cm2 C) 7 cm2 D) 14 cm2

Answer»

Correct option is  C) 7 cm2
Discussion
104.

Area of Rhombus ABCD is 27 cm2 , if the diagonal BD is 6 cm, then the length of other diagonal AC is A) 4.5 cm B) 9 cm C) 3 cm D) 12 cm

Answer»

Correct option is  B) 9 cm
Discussion