In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.5 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.5 pdf, free KSEEB SSLC Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.5 pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.5**

Question 1.

In ∆ ABC, AC = AB and the altitude AD bisects BC prove that ∆ ADC = ∆ ADB.

Answer:

In ∆ ABD and ∆ ACD

AB = AC [data]

BD = DC [data]

AD = AD [Common side]

∴ ∆ ABD = ∆ ACD [SSS postulate].

Question 2.

In a square PQRS, diagonals bisect each other at O. Prove that: ∆ POQ = ∆ QOR = ∆ ROS = ∆ SOP

Answer:

∆ POQ and ∆ QOR

PQ = QR [Sides of a square]

PO = OR [data]

OQ = OQ [Common side]

∴ ∆ POQ = ∆ QOR [SSS postulate] …i In AQOR and AROS QR = RS [Sides of a square]

QO = OS [data]

OR = OR [Common side]

∴ ∆ QOR = ∆ ROS [SSS postulate] …ii

In ∆ROS and ∆SOP

SR = SP [Sides of a square]

RO = OP [data]

OS = OS [Common side]

∴ ∆ ROS ≅ ∆SOP [SSSpostulate]…iii

From (i), (ii) and (iii)

∆POQ ≅ ∆QOR ≅ ∆ROS ≅ ∆SOP

Question 3.

In figure two sides AB, BC, and the median AD of ∆ABC are respectively equal to two sides PQ, QR, and median PS of ∆PQR. Prove that

(1) AADB ≅ APSQ

(2) AADC ≅ APSR

Does if follow that triangles ABC and PQR are congruent.

Ans

1. IN ∆ ADB and ∆ PSQ

AB = PQ [data]

BD = QS [BC = QR, D and S are the mid points]

AD = PS [data]

∆ ABD ≅ ∆PQR [SSS postulate]

∠APB = ∠ PSQ [Corresponding sides]

2. ∆ ADC and ∆ PSR

AB = PQ [data]

DC = SR

[BC = QR, D and S are the mid points]

∠ADC =∠PSR

∴[∠ADC = 180 – ∠APB

∠PSR = 180 -∠PSQ

∠ADB =∠PSQ]

∆ ABC ≅ ∆ PSR [SAS postulate]

Yes from the results of (i) and( ii)

∆ ABC = ∆ PQR

Question 4.

In ∆ PQR, PQ = QR, L, M, and N are the midpoint of the sides of PQ, QR, and RP respectively prove that LN = MN

Answer:

PQ = QR [data]

∠P = ∠R [Theorem l]

In ∆ PLN and RMN

PL = RM

[PQ = QR and L and M are the midpoints]

∠P = ∠R

PN = NR [N is the mid point of PR]

∴ ∆ PLN ≅ ∆RMN [SAS postulate]

∴ LN = MN [Corresponding sides]

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