Advertisements
Advertisements
प्रश्न
In the below Fig, ABC and ABD are two triangles on the base AB. If line segment CD is
bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD)

Advertisements
उत्तर
Given that CD is bisected at O by AB
To prove: ar (ΔABC) = ar (ΔABD)
Construction: Draw CP ⊥ AB and DQ ⊥ AB
Proof:-
`ar (ΔABC) = 1/2 xx AB xx CP` ........ (1)
`ar (ΔABC) = 1/2 xx AB xx DQ ` ........ (2)
In ∠CPO and ΔDQO
∠CPQ = ΔDQO [Each 90°]
Given that CO = DO
∠COP = ∠DOQ [vertically opposite angles are equal]
Than, ΔCPO ≅ DQO [By AAS condition]
∴ CP = DQ ........... (3) [CP.C.T]
Compare equation (1), (2) and (3)
Area ( ΔABC)a = area of ΔABD
APPEARS IN
संबंधित प्रश्न
In the below fig. ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm,
and distance between AB and DC is 4cm. Find the value of x and area of trapezium ABCD.

A, B, C, D are mid-points of sides of parallelogram PQRS. If ar (PQRS) = 36 cm2, then ar (ABCD) =
If AD is median of ΔABC and P is a point on AC such that
ar (ΔADP) : ar (ΔABD) = 2 : 3, then ar (Δ PDC) : ar (Δ ABC)
ABCD is a trapezium in which AB || DC. If ar (ΔABD) = 24 cm2 and AB = 8 cm, then height of ΔABC is
Find the area of a rectangle whose length and breadth are 25 m and 16 cm.
Find the area of a rectangle whose length = 3.6 m breadth = 90 cm
The King was very happy with carpenters Cheggu and Anar. They had made a very big and beautiful bed for him. So as gifts the king wanted to give some land to Cheggu, and some gold to Anar. Cheggu was happy. He took 100 meters of wire and tried to make different rectangles.
He made a 10 m × 40 m rectangle. Its area was 400 square meters. So he next made a 30 m × 20 m rectangle.
- What other rectangles can he make with 100 meters of wire? Discuss which of these rectangles will have the biggest area.
Is the area of both your footprints the same?
Find the area of the following figure by counting squares:

What is the area of a closed shape?
