Ncert Solutions For Class 9 Maths Ex 7.3

Ncert Solutions For Class 9 Maths Chapter 7 Ex 7.3

Question 1:

On the same base BC XYZ and DYZ are two isosceles triangles and vertices X and D are on the same side of YZ. If XD is extended to intersect YZ at P, show that

(i) XYD XZD

(ii) XYP XZP

(iii) XP bisects X as well as D.

(iv) XP is the perpendicular and bisector of YZ.

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Solution:

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(i) In ⧍XYD and ⧍XZD, we have

XY = XZ [Given]

YD = ZD [Given]

XD = XD [Common]

⧍XYD ⩬ ⧍XZD [SSS congruence].

(ii) In ⧍XYP and ⧍XZP, we have

XY = XZ

∠YXP = ∠ZXP

XP = XP

⧍XYP ⩬ ⧍XZP [SAS congruence].

(iii)⧍XYD ⩬ ⧍XDY

∠XDY = ∠XDZ

=> 180° — ∠XDY = 180° — ∠XDZ
Also, from part (ii), ∠YXPD = ∠ZXP

XP bisects DX as well as ∠D.

(iv) Now, YP = ZP and ∠YPX = ∠ZPX

But ∠YPX + ∠ZPX = 180°

So, 2∠BPA = 180°

Or, ∠YPX = 90°

Since YP = ZP, therefore XP is perpendicular bisector of YZ.

 

Question 2:

Isosceles triangle XYZ has an altitude XD, in which XY = XZ. Show that

(i) XD bisects YZ

(ii) XD bisects X

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Solution:

(i) In ⧍ABD and ⧍ACD, we have ∠ADB = ∠ADC

XY = XZ

XD = XD

⧍XYD ⩬ ⧍XZD.

YD = ZD

Hence, XD bisects YC.

(ii) Also, ∠YXD = ∠ZXD

Hence XD bisects ∠X.

 

Question 3:

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of PQR. Show that:

(i) ABM   APQN (ii) ABC PQR

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Solution:

(i) In ⧍ABM and ⧍PQN,

we have

BM = QN

12BC=12QR

AB = PQ

AM = PN

⧍ABM = ⧍PQN [ SSS Congruence ]

∠ABM = ∠PQN.

(ii) Now, in ⧍ABC and ⧍PQR, we have

AB = PQ

∠ABC = ∠PQR

BC = QR

⧍ABC ⩬  ⧍PQR [SAS congruence]

 

Question 4:

In triangle ABC, BE and CF are two equal altitudes. By Using RHS congruence rule, prove that the triangle ABC is isosceles.

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Solution:

BE and CF are altitudes of a ⧍ABC.

∠BEC = ∠CFB = 90°

Now, in right triangles BEB and CFB, we have

Hyp. BC = Hyp. BC

Side BE= Side CF

⧍BEC ⩬  ⧍CFB

∠BCE = ∠CBF

Now, in ⧍ABC, ∠B = ∠C

AB = AC

Hence, ⧍ABC is an isosceles triangle.

 

Question 5:

AB = AC in an isosceles triangle ABC. Draw AP perpendicular BC to show that ∠B= ∠C.

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Solution:

Draw AP perpendicular BC. In ⧍ABP and ⧍ACP, we have

AB = AC

∠APB = ∠APC

AB = AP [Common]

⧍ABP = ⧍ACP [By RHS congruence rule]

Also, ∠B = ∠C Proved. [CPCT]

 

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