Advertisements
Advertisements
प्रश्न
In parallelogram LOST, SN ⊥ OL and SM ⊥ LT. Find ∠STM, ∠SON and ∠NSM.

Advertisements
उत्तर
Given, ∠MST = 40°
In ΔMST,
By the angle sum property of a triangle,
∠TMS + ∠MST + ∠STM = 180°
⇒ ∠STM = 180° – (90° + 40°) ...[∵ SM ⊥ LT, ∠TMS = 90°]
= 50°
∴ ∠SON = ∠STM = 50° ...[∵ Opposite angles of a parallelogram are equal]
Now, In the ΔONS,
∠ONS + ∠OSN + ∠SON = 180° ...[Angle sum property of triangle]
∠OSN = 180° – (90° + 50°)
= 180° – 140°
= 40°
Moreover, ∠SON + ∠TSO = 180° ...[∵ Adjacent angles of a parallelogram are supplementary]
⇒ ∠SON + ∠TSM + ∠NSM + ∠OSN = 180°
⇒ 50° + 40° + ∠NSM + 40° = 180°
⇒ 90° + 40° + ∠NSM = 180°
⇒ 130° + ∠NSM = 180°
⇒ ∠NSM = 180° – 130° = 50°
APPEARS IN
संबंधित प्रश्न
Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side by 25 cm. Find the lengths of all sides.
In the given figure, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS then prove that `square`PQRS is a parallelogram.

Construct a parallelogram ABCD such that l(BC) = 7 cm, m∠ABC = 40° , l(AB) = 3 cm.
In parallelogram ABCD, ∠A = 3 times ∠B. Find all the angles of the parallelogram. In the same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD.
ABCD is a parallelogram. What kind of quadrilateral is it if: AC is perpendicular to BD but is not equal to it?
Prove that the diagonals of a parallelogram bisect each other.
The given figure shows parallelogram ABCD. Points M and N lie in diagonal BD such that DM = BN.

Prove that:
(i) ∆DMC = ∆BNA and so CM = AN
(ii) ∆AMD = ∆CNB and so AM CN
(iii) ANCM is a parallelogram.
ABCD is a parallelogram. Find the value of x, y and z.

ABCD is a parallelogram. Points P and Q are taken on the sides AB and AD respectively and the parallelogram PRQA is formed. If ∠C = 45°, find ∠R.
The following figure RUNS is parallelogram. Find x and y. (Lengths are in cm)

