3. Triangles

Choose the correct alternative answer for the following questions:

Click on the question to view the answer

The sum of any two sides of a triangle is greater than the third side. ... (Theorem)
So, if the third side is considered as 3.4 cm, then 3.4 + 1.5 = 4.9, which is not greater than the remaining side 5.
Hence, the third side cannot be 3.4 cm.
∴ The correct option is: (D) 3.4 cm

Solution:

If two angles of a triangle are unequal then the side opposite to the greater angle is greater than the side opposite to smaller angle. ... (Theorem)

In \(\triangle\)PQR,

\(\angle\)R > \(\angle\)Q ... (Given)

∴ PQ > PR

∴ The correct option is: (B) PQ > PR

Solution:

In \(\triangle\)TPQ,
\(\angle\)T + \(\angle\)P + \(\angle\)Q = 180\(^{\circ}\) ... (Theorem)
∴ 65\(^{\circ}\) + 95\(^{\circ}\) + \(\angle\)Q = 180\(^{\circ}\)
∴ 160\(^{\circ}\) + \(\angle\)Q = 180\(^{\circ}\)
∴ \(\angle\)Q = 180\(^{\circ}\) − 160\(^{\circ}\)
∴ \(\angle\)Q = 20\(^{\circ}\) ... (i)

If two angles of a triangle are unequal then the side opposite to the greater angle is greater than the side opposite to smaller angle. ... (Theorem)

Here, 95\(^{\circ}\) > 65\(^{\circ}\) > 20\(^{\circ}\)
∴ \(\angle\)P > \(\angle\)T > \(\angle\)Q
∴ TQ > PQ > TP
i.e. TP < PQ < TQ

∴ The correct option is: (B) PQ < TQ

2. \(\triangle\)ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.

Solution:

Given:

In \(\triangle\)ABC,
AB = AC
BD and CE are medians.

To prove:

BD = CE

Proof:

In \(\triangle\)ABC,
AB = AC

∴ \(\displaystyle \frac{1}{2}\text{AB} = \frac{1}{2}\text{AC}\)

∴ AE = AD ... (∵ BD and CE are medians) ... (i)

Now, in \(\triangle\)ADB and \(\triangle\)AEC,
seg AB \(\cong\) seg AC ... (Given)
\(\angle\)DAB \(\cong\) \(\angle\)EAC ... (Common angle)
seg AD \(\cong\) seg AE ... [From (i)]
∴ \(\triangle\)ADB \(\cong\) \(\triangle\)AEC ... (SAS test)
∴ seg BD \(\cong\) seg CE ... (c s c t)
i.e. BE = CE

3. In \(\triangle\)PQR, If PQ > PR and bisectors of \(\angle\)Q and \(\angle\)R intersect at S. Show that SQ > SR.

Proof:

In \(\triangle\)PQR,
PQ > PR ... (Given)
∴ \(\angle\)PRQ > \(\angle\)PQR ... (Theorem)

∴ \(\displaystyle \frac{1}{2} \angle \text{PRQ} > \frac{1}{2} \angle \text{PQR}\)

∴ \(\angle\)SRQ > \(\angle\)SQR
... (∵ QS and RS are bisectors of \(\angle\)Q and \(\angle\)R respectively) ... (Given)

Now, in \(\triangle\)SQR,
SQ > SR ... (Theorem: Side opposite to greater angle is greater)

4. In figure 3.59, points D and E are on side BC of \(\triangle\)ABC, such that BD = CE and AD = AE. Show that \(\triangle\)ABD \(\cong\) \(\triangle\)ACE.

Proof:

In \(\triangle\)ADE,
seg AD \(\cong\) seg AE ... (Given)
∴ \(\angle\)ADE \(\cong\) \(\angle\)AED ... (Theorem of Isosceles Triangle) ... (i)

Now, \(\angle\)ADE + \(\angle\)ADB = 180\(^\circ\) ... (Angles in linear pair) ... (ii)

Similarly, \(\angle\)AED + \(\angle\)AEC = 180\(^\circ\) ... (Angles in linear pair) ... (iii)

From (ii) and (iii), we have:
\(\angle\)ADE + \(\angle\)ADB = \(\angle\)AED + \(\angle\)AEC
∴ \(\angle\)ADB \(\cong\) \(\angle\)AEC ... [From (i)] ... (iv)

Now, in \(\triangle\)ABD and \(\triangle\)ACE,
seg AD \(\cong\) seg AE ... (Given)
\(\angle\)ADB \(\cong\) \(\angle\)AEC ... [From (iv)]
seg BD \(\cong\) seg CE ... (Given)
∴ \(\triangle\)ABD \(\cong\) \(\triangle\)ACE ... (S A S test) ... (v)

5. In figure 3.60, point S is any point on side QR of \(\triangle\)PQR. Prove that: PQ + QR + RP > 2PS

Proof:

In \(\triangle\)PQS,
PQ + QS > PS ... (Theorem) ... (i)

Also, in \(\triangle\)PRS,
SR + RP > PS ... (Theorem) ... (ii)

Adding (i) and (ii), we get:
PQ + QS + SR + RP > PS + PS
∴ PQ + QS + SR + RP > 2PS
∴ PQ + QR + RP > 2PS ... (∵ QS + SR = QR)

6. In figure 3.61, bisector of \(\angle\)BAC intersects side BC at point D. Prove that AB > BD.

Proof:

Seg AD is the bisector of \(\angle\)BAC. ... (Given)
∴ \(\angle\)BAD \(\cong\) \(\angle\)DAC ... (i)

Now, \(\angle\)ADB is an exterior angle of \(\triangle\)ADC.
∴ \(\angle\)ADB > \(\angle\)DAC ... (Exterior angle theorem) ... (ii)
i.e. \(\angle\)ADB > \(\angle\)BAD ... [From (i)] ... (ii)

∴ In \(\triangle\)ABD,
seg AB > seg BD ... (Converse of Exterior Angle Theorem)
i.e. AB > BD ... (iii)

7. In figure 3.62, seg PT is the bisector of \(\angle\)QPR. A line parallel to seg PT and passing through R intersects ray QP at point S. Prove that PS = PR.

Proof:

Seg PT is the bisector of \(\angle\)QPR. ... (Given)
∴ \(\angle\)QPT \(\cong\) \(\angle\)RPT ... (i)

Also, seg PT | | seg SR. ... (Given)
Consider transversal QS.
∴ \(\angle\)QPT \(\cong\) \(\angle\)PSR ... (Corresponding angles) ... (ii)
Consider transversal PR.
∴ \(\angle\)RPT \(\cong\) \(\angle\)PRS ... (Alternate interior angles) ... (iii)

From (i), (ii), and (iii), we have:
\(\angle\)PSR \(\cong\) \(\angle\)PRS ... (iv)

∴ In \(\triangle\)RPS,
∴ PS = PR ... (Converse of Theorem of Isosceles Triangle) ... (v)

8. In figure 3.63, seg AD \(\perp\) seg BC. Seg AE is the bisector of \(\angle\)CAB and
B – D – E.
Prove that: \(\displaystyle \angle \text{DAE} = \frac{1}{2}( \angle \text{B} - \angle \text{C})\)

Proof:

Seg AE is the bisector of \(\angle\)CAB. ... (Given)
∴ \(\angle\)BAE \(\cong\) \(\angle\)CAE ... (i)

Now, \(\angle\)AEB is an exterior angle of \(\triangle\)AEC.
∴ \(\angle\)AEB = \(\angle\)CAE + \(\angle\)C ... (Remote Interior Angle theorem)
∴ \(\angle\)AEB − \(\angle\)CAE = \(\angle\)C
i.e. \(\angle\)C = \(\angle\)AEB − \(\angle\)CAE ... (ii)

Also, in \(\triangle\)BEA,
\(\angle\)B + \(\angle\)BAE + \(\angle\)AEB = 180° ... (Theorem)
∴ \(\angle\)B = 180° − \(\angle\)BAE − \(\angle\)AEB ... (iii)

Subtracting (ii) from (iii):
\(\angle\)B − \(\angle\)C = [180° − \(\angle\)BAE − \(\angle\)AEB] − [\(\angle\)AEB − \(\angle\)CAE]
∴ \(\angle\)B − \(\angle\)C = 180° − \(\angle\)BAE − \(\angle\)AEB − \(\angle\)AEB + \(\angle\)CAE
∴ \(\angle\)B − \(\angle\)C = 180° − \(\cancel{\angle \text{BAE}}\) − \(\angle\)AEB − \(\angle\)AEB + \(\cancel{\angle \text{CAE}}\) ... [From (i)]
∴ \(\angle\)B − \(\angle\)C = 180° − 2\(\angle\)AEB ... (iv)

Now, in \(\triangle\)ADE,
\(\angle\)DAE + \(\angle\)ADE + \(\angle\)AED = 180 ... (Theorem)
∴ \(\angle\)DAE + 90° + \(\angle\)AED = 180° ... [∵ \(\angle\)ADE = 90° ... (Given)]
∴ \(\angle\)DAE = 180° − 90° − \(\angle\)AED
∴ \(\angle\)DAE = 90° − \(\angle\)AED

∴ \(\displaystyle \angle \text{DAE} = \frac{1}{2}( 180° - 2\angle \text{AED})\)

∴ \(\displaystyle \angle \text{DAE} = \frac{1}{2}( 180° - 2\angle \text{AEB})\) ... (v)

From (iv) and (v),

\(\displaystyle \angle \text{DAE} = \frac{1}{2}( \angle \text{B} − \angle \text{C})\)

Choose the correct alternative answer for the following questions: Click on the question to view the answer

If two sides of a triangle are 5 cm and 1.5 cm, the lenght of its third side cannot be _____. 3.7 cm 4.1 cm 3.8 cm 3.4 cm

In \(\triangle\)PQR, if \(\angle\)R > \(\angle\)Q then _____. QR > PR PQ > PR PQ < PR QR < PR

Solution:

In \(\triangle\)TPQ, \(\angle\)T = 65\(^{\circ}\), \(\angle\)P = 95\(^{\circ}\) then, which of the following statements is true? PQ < TP PQ < TQ TQ < TP < PQ PQ < TP < TQ

Solution:

2. \(\triangle\)ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.

Solution:

Given:

To prove:

Proof:

3. In \(\triangle\)PQR, If PQ > PR and bisectors of \(\angle\)Q and \(\angle\)R intersect at S. Show that SQ > SR.

Proof:

4. In figure 3.59, points D and E are on side BC of \(\triangle\)ABC, such that BD = CE and AD = AE. Show that \(\triangle\)ABD \(\cong\) \(\triangle\)ACE.

Proof:

5. In figure 3.60, point S is any point on side QR of \(\triangle\)PQR. Prove that: PQ + QR + RP > 2PS

Proof:

6. In figure 3.61, bisector of \(\angle\)BAC intersects side BC at point D. Prove that AB > BD.

Proof:

7. In figure 3.62, seg PT is the bisector of \(\angle\)QPR. A line parallel to seg PT and passing through R intersects ray QP at point S. Prove that PS = PR.

Proof:

8. In figure 3.63, seg AD \(\perp\) seg BC. Seg AE is the bisector of \(\angle\)CAB andB – D – E. Prove that: \(\displaystyle \angle \text{DAE} = \frac{1}{2}( \angle \text{B} - \angle \text{C})\)

Proof:

Choose the correct alternative answer for the following questions:

Click on the question to view the answer

8. In figure 3.63, seg AD \(\perp\) seg BC. Seg AE is the bisector of \(\angle\)CAB and
B – D – E.
Prove that: \(\displaystyle \angle \text{DAE} = \frac{1}{2}( \angle \text{B} - \angle \text{C})\)