Advertisements
Advertisements
प्रश्न
PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and B is a point on SR such that AB || QR. If area of ΔPBQ is 17cm2, find the area of ΔASR.
Advertisements
उत्तर
Given: Here from the given figure we get
(1) PQRS is a trapezium having PS||QR
(2) A is any point on PQ
(3) B is any point on SR
(4) AB||QR
(5) Area of ΔBPQ = 17 cm2
To find : Area of ΔASR

Calculation: We know that ‘If a triangle and a parallelogram are on the same base and the same parallels, the area of the triangle is equal to half the area of the parallelogram’
Here we can see that:
Area (ΔAPB) = Area (ΔABS) …… (1)
And, Area (ΔAQR) = Area (ΔABR) …… (2)
Therefore,
Area (ΔASR) = Area (ΔABS) + Area (ΔABR)
From equation (1) and (2), we have,
Area (ΔASR) = Area (ΔAPB) + Area (ΔAQR)
⇒ Area (ΔASR) = Area (ΔBPQ) = 17 cm2
Hence, the area of the triangle ΔASR is 17 cm2.
APPEARS IN
संबंधित प्रश्न
If ABCD is a parallelogram, then prove that
𝑎𝑟 (Δ𝐴𝐵𝐷) = 𝑎𝑟 (Δ𝐵𝐶𝐷) = 𝑎𝑟 (Δ𝐴𝐵𝐶) = 𝑎𝑟 (Δ𝐴𝐶𝐷) = `1/2` 𝑎𝑟 (||𝑔𝑚 𝐴𝐵𝐶𝐷) .
In below fig., PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove
that ar (Δ PQE) = ar (ΔCFD).

In a ΔABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If ar (ΔABC) = 16cm2, then ar (trapezium FBCE) =
In the given figure, PQRS is a parallelogram. If X and Y are mid-points of PQ and SRrespectively and diagonal Q is joined. The ratio ar (||gm XQRY) : ar (ΔQSR) =

ABCD is a trapezium with parallel sides AB =a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFEand EFCD is
The sides of a rectangular park are in the ratio 4 : 3. If its area is 1728 m2, find
(i) its perimeter
(ii) cost of fencing it at the rate of ₹40 per meter.
Find the area of a square, whose side is: 7.2 cm.
By counting squares, estimate the area of the figure.

Find the area of the following figure by counting squares:

When using a square centimeter sheet for measurement, the area of one entire square equals:
