CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The following figure shows a trapezium ABCD in which AB || DC. 
P is the mid-point of AD and PR || AB. Then:

194305_44cf5ad7019d488daad7467a99ea31a7.png


A
PR=12(AB+CD)
loader
B
PR=(AB+CD)
loader
C
PR=13(AB+CD)
loader
D
PR=14(AB+CD)
loader

Solution

The correct option is A $$PR = \displaystyle \frac{1}{2} (AB + CD)$$
In $$\triangle $$, BDC and BQR,
$$\angle B = \angle B$$ (Common)
$$\angle BDC = \angle BQR$$ (Corresponding angles of parallel lines)
$$\angle BCD = \angle BRQ$$ (Corresponding angles of parallel lines)
thus, $$\triangle BDC \sim \triangle BQR$$
Thus, $$\dfrac{BD}{QB} = \dfrac{DC}{QR}$$
$$2 = \dfrac{DC}{QR}$$ (Q is the mid point of BD)
$$QR = \dfrac{1}{2} DC $$
Similarly, $$QP = \dfrac{1}{2} AB $$
Hence, $$QR + QP = \dfrac{1}{2} (AB + DC)$$
$$ PR = \dfrac{1}{2}(AB + CD)$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image