Advertisements
Advertisements
प्रश्न
In the given figure, the perimeter of parallelogram PQRS is 42 cm. Find the lengths of PQ and PS.
Advertisements
उत्तर
Area of ||gm PQRS = PQ x 6
Also,
Area of ||gm PQRS = PS x 8
∴ PQ x 6 = PS x 8
⇒ PQ = `(8"PS")/(6)`
⇒ PQ = `(4"PS")/(3)` ....(i)
Perimeter of ||gm PQRS = PQ + OR + RS + PS
⇒ 42 = 2PQ + 2PS ...(opposite sides of a parallelogram are equal)
⇒ 21 = PQ + PS
⇒ `(4"PS")/(3) + "PS"` = 21 ...[From (i)]
⇒ `(4"PS" + 3"PS")/(3)`
⇒ 7PS = 63
⇒ PS = 9cm
Now,
PQ = `(4"PS")/(3)`
= `(4 xx 9)/(3)`
= 12cm
∴ PQ = 12cm and PS = 9cm.
APPEARS IN
संबंधित प्रश्न
E is the mid-point of side AB and F is the mid-point of side DC of parallelogram ABCD. Prove that AEFD is a parallelogram.
The alongside figure shows a parallelogram ABCD in which AE = EF = FC.
Prove that:
- DE is parallel to FB
- DE = FB
- DEBF is a parallelogram.

In the given figure, ABCD is a parallelogram.
Prove that: AB = 2 BC.

The following figure shows a trapezium ABCD in which AB is parallel to DC and AD = BC. 
Prove that:
(i) ∠DAB = ∠CBA
(ii) ∠ADC = ∠BCD
(iii) AC = BD
(iv) OA = OB and OC = OD.
In parallelogram ABCD, the bisector of angle A meets DC at P and AB = 2 AD.
Prove that:
(i) BP bisects angle B.
(ii) Angle APB = 90o.
Points M and N are taken on the diagonal AC of a parallelogram ABCD such that AM = CN. Prove that BMDN is a parallelogram.
PQRS is a parallelogram. T is the mid-point of PQ and ST bisects ∠PSR.
Prove that: QR = QT
ABCD is a parallelogram. The bisector of ∠BAD meets DC at P, and AD is half of AB.
Prove that: ∠APB is a right angle.
In the given figure, MP is the bisector of ∠P and RN is the bisector of ∠R of parallelogram PQRS. Prove that PMRN is a parallelogram.
In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Lines DP and CQ meet at O. Show that the area of triangle QPO is `9/8` of the area of the parallelogram ABCD
