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TRIANGLES

90665 Choose the correct answer from the given four options:
In triangles ABC and DEF, $∠ B = ∠ E, ∠ F = ∠ C and AB = 3 DE .$ Then, the two triangles are:

1 Congruent but not similar.

2 Similar but not congruent.

3 Neither congruent nor similar.

4 Congruent as well as similar.

Explanation:

BSimilar but not congruent.
$In △ ABC and △ DEF, ∠ B = ∠ E = ∠ F and = AB = 3 DE$

We know that, if in two triangles corresponding two angles are same, then they are similar by AAA similarity criterion. Also $△ ABC and △ DEF$ do not satisfy any rule of congruency, (SAS, ASA, SSS), so both are not congruent.
Understanding

TRIANGLES

90666 A man goes 24m due west and then 10m due north. How far is he from the starting point?

1 34m

2 17m

3 26m

4 28m

Explanation:

C26m

Let O be the starting point.
From O the man goes west that is towards, W till point A. He then moves 10m due nirth, that is towards N to point B.
$△ OAB$ forms a right - angled triangle.
By Pythagoras theorem,
OB $^{2}$ = OA $^{2}$ + AB $^{2}$
OB $^{2}$ = $^{}$ 24 $^{2}$ + 10 $^{2}$
OB $^{2}$ = 576 + 100
OB $^{2}$ = 676
OB = 26m
So, the man is 26m away from the the starting point.

TRIANGLES

90685 If $△ ABC \sim △ QRP, \frac{ar (△ ABC)}{ar (△ PQR)} = \frac{9}{4}, AB = 18 cm$ and $BC = 15 cm$ then $PR = ?$

1 8cm

2 10cm

3 12cm

4

\frac{20}{3} cm

Explanation:

B10cm
$△ ABC \sim △ QRP$
$\Rightarrow \frac{AB}{QR} = \frac{BC}{PR} = \frac{AC}{PQ}$
Also, $\frac{ar (△ ABC)}{ar (△ PQR)} = \frac{{AB}^{2}}{{BC}^{2}}$
$\Rightarrow \frac{9}{4} = \frac{{AB}^{2}}{{BC}^{2}}$
$\Rightarrow \frac{AB}{QR} = \frac{3}{2}$
So, $\frac{AB}{QR} = \frac{BC}{PR} \Rightarrow \frac{BC}{PR} = \frac{3}{2}$
$\Rightarrow \frac{15}{PR} = \frac{3}{2}$
$\Rightarrow PR = 10 cm .$

TRIANGLES

90667 In the given figure if $PS ∥ QR$ and $PQ ∥ SR$ and AT = AQ = 6, AS = 3, TS = 4, then:

1 x = 3, y = 4.

2 x = 1, y = 2.

3 x = 2, y = 3.

4 x = 4, y = 5.

Explanation:

Ax = 3, y = 4.
In triangles $APQ$ and $ATS$
$∠ PAQ = TAS,$ [Verticaly opposite angles] $∠ PQA = ATS,$ [Alternate angles]
$∴ △ APQ \sim △ AST$ [AA similarity]
$∴ \frac{AQ}{AT} = \frac{AP}{AS}$
$\Rightarrow \frac{6}{6} = \frac{x}{3}$
$\Rightarrow x = \frac{6 \times 3}{6} = 3$
And $\frac{AQ}{AT} = \frac{PQ}{ST}$
$\Rightarrow \frac{6}{6} = \frac{x}{4}$
$\Rightarrow y = \frac{4 \times 6}{6} = 4$
$\Rightarrow Therefore, x = 3, y = 4$

TRIANGLES

90668 In the figure, RS || DB || PQ. If CP = PD = 11cm and DR = RA = 3cm. Then the values of x and y are respectively:

1 12, 10.

2 14, 6.

3 10, 7

4 16, 8.

Explanation:

D16, 8.
the figure RS || DB || PQ
CP = PD = 11cm DR = RA = 3cm
In $△ ABD$
RS || BD and AR = RD
$RS = \frac{1}{2} BD$
$y = \frac{1}{2} x or x = 2 y$
Only 16, 8 is possible.
Understanding

TRIANGLES

90665 Choose the correct answer from the given four options:
In triangles ABC and DEF, $∠ B = ∠ E, ∠ F = ∠ C and AB = 3 DE .$ Then, the two triangles are:

1 Congruent but not similar.

2 Similar but not congruent.

3 Neither congruent nor similar.

4 Congruent as well as similar.

Explanation:

BSimilar but not congruent.
$In △ ABC and △ DEF, ∠ B = ∠ E = ∠ F and = AB = 3 DE$

We know that, if in two triangles corresponding two angles are same, then they are similar by AAA similarity criterion. Also $△ ABC and △ DEF$ do not satisfy any rule of congruency, (SAS, ASA, SSS), so both are not congruent.
Understanding

TRIANGLES

90666 A man goes 24m due west and then 10m due north. How far is he from the starting point?

1 34m

2 17m

3 26m

4 28m

Explanation:

C26m

Let O be the starting point.
From O the man goes west that is towards, W till point A. He then moves 10m due nirth, that is towards N to point B.
$△ OAB$ forms a right - angled triangle.
By Pythagoras theorem,
OB $^{2}$ = OA $^{2}$ + AB $^{2}$
OB $^{2}$ = $^{}$ 24 $^{2}$ + 10 $^{2}$
OB $^{2}$ = 576 + 100
OB $^{2}$ = 676
OB = 26m
So, the man is 26m away from the the starting point.

TRIANGLES

90685 If $△ ABC \sim △ QRP, \frac{ar (△ ABC)}{ar (△ PQR)} = \frac{9}{4}, AB = 18 cm$ and $BC = 15 cm$ then $PR = ?$

1 8cm

2 10cm

3 12cm

4

\frac{20}{3} cm

Explanation:

B10cm
$△ ABC \sim △ QRP$
$\Rightarrow \frac{AB}{QR} = \frac{BC}{PR} = \frac{AC}{PQ}$
Also, $\frac{ar (△ ABC)}{ar (△ PQR)} = \frac{{AB}^{2}}{{BC}^{2}}$
$\Rightarrow \frac{9}{4} = \frac{{AB}^{2}}{{BC}^{2}}$
$\Rightarrow \frac{AB}{QR} = \frac{3}{2}$
So, $\frac{AB}{QR} = \frac{BC}{PR} \Rightarrow \frac{BC}{PR} = \frac{3}{2}$
$\Rightarrow \frac{15}{PR} = \frac{3}{2}$
$\Rightarrow PR = 10 cm .$

TRIANGLES

90667 In the given figure if $PS ∥ QR$ and $PQ ∥ SR$ and AT = AQ = 6, AS = 3, TS = 4, then:

1 x = 3, y = 4.

2 x = 1, y = 2.

3 x = 2, y = 3.

4 x = 4, y = 5.

Explanation:

Ax = 3, y = 4.
In triangles $APQ$ and $ATS$
$∠ PAQ = TAS,$ [Verticaly opposite angles] $∠ PQA = ATS,$ [Alternate angles]
$∴ △ APQ \sim △ AST$ [AA similarity]
$∴ \frac{AQ}{AT} = \frac{AP}{AS}$
$\Rightarrow \frac{6}{6} = \frac{x}{3}$
$\Rightarrow x = \frac{6 \times 3}{6} = 3$
And $\frac{AQ}{AT} = \frac{PQ}{ST}$
$\Rightarrow \frac{6}{6} = \frac{x}{4}$
$\Rightarrow y = \frac{4 \times 6}{6} = 4$
$\Rightarrow Therefore, x = 3, y = 4$

TRIANGLES

90668 In the figure, RS || DB || PQ. If CP = PD = 11cm and DR = RA = 3cm. Then the values of x and y are respectively:

1 12, 10.

2 14, 6.

3 10, 7

4 16, 8.

Explanation:

D16, 8.
the figure RS || DB || PQ
CP = PD = 11cm DR = RA = 3cm
In $△ ABD$
RS || BD and AR = RD
$RS = \frac{1}{2} BD$
$y = \frac{1}{2} x or x = 2 y$
Only 16, 8 is possible.
Understanding

TRIANGLES

90665 Choose the correct answer from the given four options:
In triangles ABC and DEF, $∠ B = ∠ E, ∠ F = ∠ C and AB = 3 DE .$ Then, the two triangles are:

1 Congruent but not similar.

2 Similar but not congruent.

3 Neither congruent nor similar.

4 Congruent as well as similar.

Explanation:

BSimilar but not congruent.
$In △ ABC and △ DEF, ∠ B = ∠ E = ∠ F and = AB = 3 DE$

We know that, if in two triangles corresponding two angles are same, then they are similar by AAA similarity criterion. Also $△ ABC and △ DEF$ do not satisfy any rule of congruency, (SAS, ASA, SSS), so both are not congruent.
Understanding

TRIANGLES

90666 A man goes 24m due west and then 10m due north. How far is he from the starting point?

1 34m

2 17m

3 26m

4 28m

Explanation:

C26m

Let O be the starting point.
From O the man goes west that is towards, W till point A. He then moves 10m due nirth, that is towards N to point B.
$△ OAB$ forms a right - angled triangle.
By Pythagoras theorem,
OB $^{2}$ = OA $^{2}$ + AB $^{2}$
OB $^{2}$ = $^{}$ 24 $^{2}$ + 10 $^{2}$
OB $^{2}$ = 576 + 100
OB $^{2}$ = 676
OB = 26m
So, the man is 26m away from the the starting point.

TRIANGLES

90685 If $△ ABC \sim △ QRP, \frac{ar (△ ABC)}{ar (△ PQR)} = \frac{9}{4}, AB = 18 cm$ and $BC = 15 cm$ then $PR = ?$

1 8cm

2 10cm

3 12cm

4

\frac{20}{3} cm

Explanation:

B10cm
$△ ABC \sim △ QRP$
$\Rightarrow \frac{AB}{QR} = \frac{BC}{PR} = \frac{AC}{PQ}$
Also, $\frac{ar (△ ABC)}{ar (△ PQR)} = \frac{{AB}^{2}}{{BC}^{2}}$
$\Rightarrow \frac{9}{4} = \frac{{AB}^{2}}{{BC}^{2}}$
$\Rightarrow \frac{AB}{QR} = \frac{3}{2}$
So, $\frac{AB}{QR} = \frac{BC}{PR} \Rightarrow \frac{BC}{PR} = \frac{3}{2}$
$\Rightarrow \frac{15}{PR} = \frac{3}{2}$
$\Rightarrow PR = 10 cm .$

TRIANGLES

90667 In the given figure if $PS ∥ QR$ and $PQ ∥ SR$ and AT = AQ = 6, AS = 3, TS = 4, then:

1 x = 3, y = 4.

2 x = 1, y = 2.

3 x = 2, y = 3.

4 x = 4, y = 5.

Explanation:

Ax = 3, y = 4.
In triangles $APQ$ and $ATS$
$∠ PAQ = TAS,$ [Verticaly opposite angles] $∠ PQA = ATS,$ [Alternate angles]
$∴ △ APQ \sim △ AST$ [AA similarity]
$∴ \frac{AQ}{AT} = \frac{AP}{AS}$
$\Rightarrow \frac{6}{6} = \frac{x}{3}$
$\Rightarrow x = \frac{6 \times 3}{6} = 3$
And $\frac{AQ}{AT} = \frac{PQ}{ST}$
$\Rightarrow \frac{6}{6} = \frac{x}{4}$
$\Rightarrow y = \frac{4 \times 6}{6} = 4$
$\Rightarrow Therefore, x = 3, y = 4$

TRIANGLES

90668 In the figure, RS || DB || PQ. If CP = PD = 11cm and DR = RA = 3cm. Then the values of x and y are respectively:

1 12, 10.

2 14, 6.

3 10, 7

4 16, 8.

Explanation:

D16, 8.
the figure RS || DB || PQ
CP = PD = 11cm DR = RA = 3cm
In $△ ABD$
RS || BD and AR = RD
$RS = \frac{1}{2} BD$
$y = \frac{1}{2} x or x = 2 y$
Only 16, 8 is possible.
Understanding

TRIANGLES

90665 Choose the correct answer from the given four options:
In triangles ABC and DEF, $∠ B = ∠ E, ∠ F = ∠ C and AB = 3 DE .$ Then, the two triangles are:

1 Congruent but not similar.

2 Similar but not congruent.

3 Neither congruent nor similar.

4 Congruent as well as similar.

Explanation:

BSimilar but not congruent.
$In △ ABC and △ DEF, ∠ B = ∠ E = ∠ F and = AB = 3 DE$

We know that, if in two triangles corresponding two angles are same, then they are similar by AAA similarity criterion. Also $△ ABC and △ DEF$ do not satisfy any rule of congruency, (SAS, ASA, SSS), so both are not congruent.
Understanding

TRIANGLES

90666 A man goes 24m due west and then 10m due north. How far is he from the starting point?

1 34m

2 17m

3 26m

4 28m

Explanation:

C26m

Let O be the starting point.
From O the man goes west that is towards, W till point A. He then moves 10m due nirth, that is towards N to point B.
$△ OAB$ forms a right - angled triangle.
By Pythagoras theorem,
OB $^{2}$ = OA $^{2}$ + AB $^{2}$
OB $^{2}$ = $^{}$ 24 $^{2}$ + 10 $^{2}$
OB $^{2}$ = 576 + 100
OB $^{2}$ = 676
OB = 26m
So, the man is 26m away from the the starting point.

TRIANGLES

90685 If $△ ABC \sim △ QRP, \frac{ar (△ ABC)}{ar (△ PQR)} = \frac{9}{4}, AB = 18 cm$ and $BC = 15 cm$ then $PR = ?$

1 8cm

2 10cm

3 12cm

4

\frac{20}{3} cm

Explanation:

B10cm
$△ ABC \sim △ QRP$
$\Rightarrow \frac{AB}{QR} = \frac{BC}{PR} = \frac{AC}{PQ}$
Also, $\frac{ar (△ ABC)}{ar (△ PQR)} = \frac{{AB}^{2}}{{BC}^{2}}$
$\Rightarrow \frac{9}{4} = \frac{{AB}^{2}}{{BC}^{2}}$
$\Rightarrow \frac{AB}{QR} = \frac{3}{2}$
So, $\frac{AB}{QR} = \frac{BC}{PR} \Rightarrow \frac{BC}{PR} = \frac{3}{2}$
$\Rightarrow \frac{15}{PR} = \frac{3}{2}$
$\Rightarrow PR = 10 cm .$

TRIANGLES

90667 In the given figure if $PS ∥ QR$ and $PQ ∥ SR$ and AT = AQ = 6, AS = 3, TS = 4, then:

1 x = 3, y = 4.

2 x = 1, y = 2.

3 x = 2, y = 3.

4 x = 4, y = 5.

Explanation:

Ax = 3, y = 4.
In triangles $APQ$ and $ATS$
$∠ PAQ = TAS,$ [Verticaly opposite angles] $∠ PQA = ATS,$ [Alternate angles]
$∴ △ APQ \sim △ AST$ [AA similarity]
$∴ \frac{AQ}{AT} = \frac{AP}{AS}$
$\Rightarrow \frac{6}{6} = \frac{x}{3}$
$\Rightarrow x = \frac{6 \times 3}{6} = 3$
And $\frac{AQ}{AT} = \frac{PQ}{ST}$
$\Rightarrow \frac{6}{6} = \frac{x}{4}$
$\Rightarrow y = \frac{4 \times 6}{6} = 4$
$\Rightarrow Therefore, x = 3, y = 4$

TRIANGLES

90668 In the figure, RS || DB || PQ. If CP = PD = 11cm and DR = RA = 3cm. Then the values of x and y are respectively:

1 12, 10.

2 14, 6.

3 10, 7

4 16, 8.

Explanation:

D16, 8.
the figure RS || DB || PQ
CP = PD = 11cm DR = RA = 3cm
In $△ ABD$
RS || BD and AR = RD
$RS = \frac{1}{2} BD$
$y = \frac{1}{2} x or x = 2 y$
Only 16, 8 is possible.
Understanding

TRIANGLES

90665 Choose the correct answer from the given four options:
In triangles ABC and DEF, $∠ B = ∠ E, ∠ F = ∠ C and AB = 3 DE .$ Then, the two triangles are:

1 Congruent but not similar.

2 Similar but not congruent.

3 Neither congruent nor similar.

4 Congruent as well as similar.

Explanation:

BSimilar but not congruent.
$In △ ABC and △ DEF, ∠ B = ∠ E = ∠ F and = AB = 3 DE$

We know that, if in two triangles corresponding two angles are same, then they are similar by AAA similarity criterion. Also $△ ABC and △ DEF$ do not satisfy any rule of congruency, (SAS, ASA, SSS), so both are not congruent.
Understanding

TRIANGLES

90666 A man goes 24m due west and then 10m due north. How far is he from the starting point?

1 34m

2 17m

3 26m

4 28m

Explanation:

C26m

Let O be the starting point.
From O the man goes west that is towards, W till point A. He then moves 10m due nirth, that is towards N to point B.
$△ OAB$ forms a right - angled triangle.
By Pythagoras theorem,
OB $^{2}$ = OA $^{2}$ + AB $^{2}$
OB $^{2}$ = $^{}$ 24 $^{2}$ + 10 $^{2}$
OB $^{2}$ = 576 + 100
OB $^{2}$ = 676
OB = 26m
So, the man is 26m away from the the starting point.

TRIANGLES

90685 If $△ ABC \sim △ QRP, \frac{ar (△ ABC)}{ar (△ PQR)} = \frac{9}{4}, AB = 18 cm$ and $BC = 15 cm$ then $PR = ?$

1 8cm

2 10cm

3 12cm

4

\frac{20}{3} cm

Explanation:

B10cm
$△ ABC \sim △ QRP$
$\Rightarrow \frac{AB}{QR} = \frac{BC}{PR} = \frac{AC}{PQ}$
Also, $\frac{ar (△ ABC)}{ar (△ PQR)} = \frac{{AB}^{2}}{{BC}^{2}}$
$\Rightarrow \frac{9}{4} = \frac{{AB}^{2}}{{BC}^{2}}$
$\Rightarrow \frac{AB}{QR} = \frac{3}{2}$
So, $\frac{AB}{QR} = \frac{BC}{PR} \Rightarrow \frac{BC}{PR} = \frac{3}{2}$
$\Rightarrow \frac{15}{PR} = \frac{3}{2}$
$\Rightarrow PR = 10 cm .$

TRIANGLES

90667 In the given figure if $PS ∥ QR$ and $PQ ∥ SR$ and AT = AQ = 6, AS = 3, TS = 4, then:

1 x = 3, y = 4.

2 x = 1, y = 2.

3 x = 2, y = 3.

4 x = 4, y = 5.

Explanation:

Ax = 3, y = 4.
In triangles $APQ$ and $ATS$
$∠ PAQ = TAS,$ [Verticaly opposite angles] $∠ PQA = ATS,$ [Alternate angles]
$∴ △ APQ \sim △ AST$ [AA similarity]
$∴ \frac{AQ}{AT} = \frac{AP}{AS}$
$\Rightarrow \frac{6}{6} = \frac{x}{3}$
$\Rightarrow x = \frac{6 \times 3}{6} = 3$
And $\frac{AQ}{AT} = \frac{PQ}{ST}$
$\Rightarrow \frac{6}{6} = \frac{x}{4}$
$\Rightarrow y = \frac{4 \times 6}{6} = 4$
$\Rightarrow Therefore, x = 3, y = 4$

TRIANGLES

90668 In the figure, RS || DB || PQ. If CP = PD = 11cm and DR = RA = 3cm. Then the values of x and y are respectively:

1 12, 10.

2 14, 6.

3 10, 7

4 16, 8.

Explanation:

D16, 8.
the figure RS || DB || PQ
CP = PD = 11cm DR = RA = 3cm
In $△ ABD$
RS || BD and AR = RD
$RS = \frac{1}{2} BD$
$y = \frac{1}{2} x or x = 2 y$
Only 16, 8 is possible.
Understanding