# NCERT Exemplar Solution for CBSE Class 10 Mathematics: Triangles (Part-IIIA)

Here you will get the CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IIIA). This part contains solutions to Q. No. 1-7 from Exercise 6.3 that consists only of the Short Answer Type Questions. These questions will prove to be very helpful while preparing for CBSE Class 10 Board Exam 2017-2018.

CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IIIA) is available here. This part of the chapter includes solutions for question number 1-8from Exercise 6.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Triangles. This exercise comprises only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

**Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Triangles:**

**Exercise 6.3**

**Short Answer Type Questions (Q. NO. 1-8):**

**Question. 1 **In a Δ*PQR, PR*^{2}* **-** PQ*^{2}* = QR*^{2}* *and *M *is a point on side *PR *such that *QM *⏊* PR. *Prove that *QM*^{2} = *PM *× *MR.*

**Solution. **

**Given: ** ln Δ*PQR*,

*PR ^{2}* -

*PQ*=

^{2}*QR*

^{2}And *QM *⏊* PR*

**To prove: ***QM*^{2}* =PM × MR*

**Proof: **Since, *PR ^{2}* -

*PQ*=

^{2}*QR*

^{2}⟹ *PR*^{2} = *PQ*^{2}* *+ *QR*^{2}

⟹ Δ*PQA* is right angled triangle at *Q* with PR being its hypotenuse.

Also, *QM *⏊* PR*

⟹ ∠QMR = ∠QMP = 90^{o}

**Question. 2 **Find the value of *x *for which *ΔEAB *in given figure.

Hence, the required value of *x* is 2

**Question. 3 **In figure, if ∠l = ∠2 and *ΔNSQ*≅* ΔMTR, *then prove that

*Δ*

*PTS*~

*Δ*

*PRQ.*

**Hence proved.**

**Question. 4 **Diagonals of a trapezium *PQRS *intersect each other at the point *O, PQ *|| *RS *and *PQ *= 3*RS*. Find the ratio of the areas of *Δ**POQ *and *Δ**ROS*.

**Solution.**

Hence, the required ratio is 9:1

**Question. 5 **In figure, if *AB *|| *DC *and *AC, PQ *intersect each other at the point O. Prove that *OA.**CQ = OC.AP. *

* *

**Hence proved.**

**Question. 6 **Find the altitude of an equilateral triangle of side 8cm.

**Solution.**

Let *ABC* be an equilateral triangle.

**Question. 7 **If *Δ**ABC *~ *Δ**DEF, AB = *4cm, *DE = *6, *EF = *9cm and *FD = *12cm, then find the perimeter of *Δ*ABC.

**Solution.**

Given, *Δ**ABC**~** **Δ**DEF*

* AB = *4cm, *D**E* = 6cm

And *EF *= 9cm, *FD = *12cm.

Thus, perimeter of D *ABC = AB* +* BC* + *AC *= 4 + 6 + 8 = 18cm

**Question. 8 **In figure, if *DE *|| *BC, *then find the ratio of ar (*Δ**AD**E*) and ar *(DECB).*

**Solution.**

Given, in *Δ**ABC*

* DE *|| *BC*

* DE *= 6cm and *BC *=12cm

Now, in *Δ**ABC *and *Δ**ADE,*

∠*ABC = *∠*ADE *[Corresponding angles]

∠*ACB *= *∠AED *[Corresponding angles]

∴ *Δ**ABC *~ *Δ**AED *[By AA** **similarity criterion]

Using the property, “The ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides,” we have:

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