Question

In ∆ABC, seg AP is an altitude on the base BC. / ∆ABC मध्ये, रेख AP हा पाया BC वरील शिरोलंब आहे. *
Captionless Image
True / बरोबर
False/चूक​

Answer 1

Answer:

Given:-

ABC is an isosceles △.

AB=AC

AD is altitude.

∠ADB=∠ADC=90°

To prove:-

(i) AD bisects BC, i.e., BD=CD

(ii) AD bisects ∠A, i.e., ∠BAD=∠CAD

Proof:-

In △ADB and △ADC,

∠ADB=∠ADC[Each 90°]

AB=AC[Given]

AD=AD[Common]

By R.H.S congruency,

△ADB≅△ADC

By C.P.C.T.

BD=CD

∠BAC=∠CAD

Hence proved.

Answer 2

Complete Question :- In ∆ABC, seg AP is an altitude on the base BC. seg BQ is an altitude on side AC , B-P-C, A-Q-C .

To Show and Find :-

• ∆CPA ~ ∆CQB .
• if AP = 7 , BQ = 8 , BC = 12 then find AC .

Solution :-

In ∆CPA and ∆CQB we have,

→ ∠CPA = ∠CQB { 90° }

→ ∠ACP = ∠BCQ { common }

So,

→ ∆CPA ~ ∆CQB { By AA similarity . }

then,

→ AP/BQ = AC/BC { when two ∆'s are similar, their corresponding sides are in same proportion .}

putting values now,

→ 7/8 = AC/12

→ 8•AC = 12 * 7

→ AC = (84/8)

→ AC = 10.5 (Ans.)

therefore, AC is equal to 10.5 .

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