- Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Let ABC is aright angled triangle, right angle at B i.e, ∟B = 90˚
Now, ∠A + ∠C = 90˚
Since the sum of ∠A and ∠B is equal to 90˚, then each ∠A and ∠B is less than 90˚
Therefore, ∠B > ∠A ⇒ AC > BC ……………..(1)
and ∠B > ∠C ⇒ AC > AB………………….(2)
Hence, (1) and (2), we have,
AC is greater than both BC and AB.
i.e, hypotenuse is greater than other two sides of a right angled triangle.
⇒ hypotenuse is the longest side of a right angled triangle.
In Fig. 7.48 sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB
Solution:
∠ABC + ∠PBC = 180˚ (linear pair)
⇒∠ABC = 180˚ – ∠PBC ……………(1)
Similarly, ∠ACB = 180˚ – ∠QCB ……………..(2)
It is given that, ∠PBC < ∠QCB
⸫ 180˚ – ∠QCB < 180˚ – ∠PBC
i.e, ∠ACB < ∠ABC [from(1) and (2)]
⇒AB < AC
Hence proved.
In the fig7.49. ∠B < ∠A and ∠C < ∠D. Show that AD < BC
Solution:
Given, ∠B < ∠A
i.e, OA < OB ………………………(1) [sides opposite to greater angle is greater]
It is also given that, ∠C < ∠D
i.e, OD < OC……………………….(2) [sides opposite to greater angle is greater]
Adding (i) and (ii),
OA + OD < OB + OC
AD < BC
- AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D
Solution:
Given, ABCD is a quadrilateral, AB is the smallest and CD is the longest side of the quadrilateral ABCD.
We have to show, ∠A > ∠C and ∠B > ∠D
Construction: Draw a line joining A and C .
In ∆ABC , AB < BC [AB is the shortest side & angle opposite to longer side is greater]
∠ACB < ∠BAC ………….(1)
In ∆ADC, AD < CD[CD is the longest side & angle opposite to longer side is greater]
∠ACD < ∠CAD …………….(2)
From (1) and (2),
∠ACB + ∠ACD < ∠BAC + ∠CAD
∠BCD < ∠BAD
Similarly, we can draw a line joining B and D to prove, ∠ADC < ∠ABC.
-
In fig PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ
Solution:
Given, PR > PQ and PS bisects ∠QPR
We have show that, ∠PSR > ∠PSQ
Since, PR > PQ
⇒∠PQR > ∠PRQ …………….(1) [angles opposite to the greater line the greater angle]
Since PS bisects ∠QPR, we have,
∠QPS = ∠RPS ……………..(2)
∠PSR is the exterior angle of ∆PQS, then,
∠PSR = ∠PQR + ∠QPS ………………..(3)
∠PSQ is the exterior angle of ∆PRS, then,
∠PSQ = ∠SPR + ∠PRQ ……………(4)
Adding (1) and (2),
∠PQR + ∠QPS > ∠PRQ + ∠RPS………………..(5)
⇒∠PSR > ∠PSQ [from (5), (3) and (4)]
- Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:
Let the point be O. Line segments from O are OP, OQ, OR and OS, where OP is perpendicular line segment l.
We have to prove, OP is the shortest l.
i.e., OP<OQ , OP < OR and OP < OS
In ∆OPQ, ∠P = 90˚
⸫ ∠Q is an acute angle (i.e, ∠Q = 90˚)
⸫ ∠Q < ∠P
Hence, OP < OQ [side opposite to greater angle is longer]
Similarly, we can prove that OP is shorter than OR , OS.
1 thought on “Triangles – Exercise 7.4 – Class IX”
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