298481
If the measures of the angles of a triangle are (2x)º , (3x - 5)º and (4x -13)º. Then the value of x is:
1 22
2 18
3 20
4 30
Explanation:
22 (2x)º + (3x - 5)º + (4x - 13)º = 180º [Angle sum property of triangle] 2x + 3x - 5 + 4x - 13 = 180º 9x - 18 = 180 9x = 198 x = 22 Hence, the correct answer is option (a).
THE TRIANGLE AND ITS PROPERTIES
298482
In Fig. if AB || CE, then the values of x and y are: 7
1 x = 26, y = 144
2 x = 36, y = 154
3 x = 154, y = 36
4 x = 144, y = 26
5 None of these.
Explanation:
None of these. \(\angle \text{DCA}+\angle \text{ACB}=180^\circ\) [Linear angles] \(\Rightarrow 134^\circ+\angle \text{ACB}=180^\circ\) \(\Rightarrow \angle \text{ACB}=46^\circ\) Now, In \(\triangle \text{ABC}\) \(\angle \text{BAC}+\angle \text{ACB}+\angle \text{ABC}=180^\circ\) [Angle sum property of triangle] \(\Rightarrow 62^\circ+46^\circ+2\text{x}^\circ=180^\circ\) \(\Rightarrow 2\text{x}^\circ=72^\circ\) \(\Rightarrow \text{x}=36\) Since, AB || CE \(\therefore \angle \text{ECB}=\angle \text{CBA}=(2\text{x})^\circ\) [Alternate angles] \(=(2\times 36)^\circ=72^\circ\) Now, \(\angle \text{DCE}+\angle \text{ECB}=180^\circ\) [Linear angles] \(\Rightarrow \text{y}^\circ+72^\circ=180^\circ\) \(\Rightarrow \text{y}=108\) Disclaimer: No Option is correct.
THE TRIANGLE AND ITS PROPERTIES
298483
Two angles of a triangle measure 90º and 30º. The measure of the third angle is.
1 90º
2 30º
3 60º
4 120º
Explanation:
60º Third angle = 180º - (90º + 30º) = 60º.
THE TRIANGLE AND ITS PROPERTIES
298484
Vikash wants to plant a flower on the ground in the form of a rhombus. The diagonals of the rhombus measure 42cm and 56cm. Find the perimeter of the field.
1 150cm
2 140cm
3 130cm
4 120cm
Explanation:
140cm Since diagonals of a rhombus bisect each other at 90° 8 Given, BD = 42cm and AC = 56cm Since BK \(=\frac{1}{2}\text{BD}=\frac{42}{2}=21\text{cm}\) Thus AK \(=\frac{1}{2}\text{AC}=\frac{56}{2}=28\text{cm}\) In \(\triangle\text{KAB}\) We have AB\(^{1}\) = AK\(^{1}\) + BK\(^{1}\) = (28)\(^{1}\) + (21)\(^{1}\) = 784 + 441 = 1225 Thus AB \(=\sqrt{1225}=35\text{cm}\) Therefore, perimeter of the field ABCD = 4 × 35 = 140cm
298481
If the measures of the angles of a triangle are (2x)º , (3x - 5)º and (4x -13)º. Then the value of x is:
1 22
2 18
3 20
4 30
Explanation:
22 (2x)º + (3x - 5)º + (4x - 13)º = 180º [Angle sum property of triangle] 2x + 3x - 5 + 4x - 13 = 180º 9x - 18 = 180 9x = 198 x = 22 Hence, the correct answer is option (a).
THE TRIANGLE AND ITS PROPERTIES
298482
In Fig. if AB || CE, then the values of x and y are: 7
1 x = 26, y = 144
2 x = 36, y = 154
3 x = 154, y = 36
4 x = 144, y = 26
5 None of these.
Explanation:
None of these. \(\angle \text{DCA}+\angle \text{ACB}=180^\circ\) [Linear angles] \(\Rightarrow 134^\circ+\angle \text{ACB}=180^\circ\) \(\Rightarrow \angle \text{ACB}=46^\circ\) Now, In \(\triangle \text{ABC}\) \(\angle \text{BAC}+\angle \text{ACB}+\angle \text{ABC}=180^\circ\) [Angle sum property of triangle] \(\Rightarrow 62^\circ+46^\circ+2\text{x}^\circ=180^\circ\) \(\Rightarrow 2\text{x}^\circ=72^\circ\) \(\Rightarrow \text{x}=36\) Since, AB || CE \(\therefore \angle \text{ECB}=\angle \text{CBA}=(2\text{x})^\circ\) [Alternate angles] \(=(2\times 36)^\circ=72^\circ\) Now, \(\angle \text{DCE}+\angle \text{ECB}=180^\circ\) [Linear angles] \(\Rightarrow \text{y}^\circ+72^\circ=180^\circ\) \(\Rightarrow \text{y}=108\) Disclaimer: No Option is correct.
THE TRIANGLE AND ITS PROPERTIES
298483
Two angles of a triangle measure 90º and 30º. The measure of the third angle is.
1 90º
2 30º
3 60º
4 120º
Explanation:
60º Third angle = 180º - (90º + 30º) = 60º.
THE TRIANGLE AND ITS PROPERTIES
298484
Vikash wants to plant a flower on the ground in the form of a rhombus. The diagonals of the rhombus measure 42cm and 56cm. Find the perimeter of the field.
1 150cm
2 140cm
3 130cm
4 120cm
Explanation:
140cm Since diagonals of a rhombus bisect each other at 90° 8 Given, BD = 42cm and AC = 56cm Since BK \(=\frac{1}{2}\text{BD}=\frac{42}{2}=21\text{cm}\) Thus AK \(=\frac{1}{2}\text{AC}=\frac{56}{2}=28\text{cm}\) In \(\triangle\text{KAB}\) We have AB\(^{1}\) = AK\(^{1}\) + BK\(^{1}\) = (28)\(^{1}\) + (21)\(^{1}\) = 784 + 441 = 1225 Thus AB \(=\sqrt{1225}=35\text{cm}\) Therefore, perimeter of the field ABCD = 4 × 35 = 140cm
298481
If the measures of the angles of a triangle are (2x)º , (3x - 5)º and (4x -13)º. Then the value of x is:
1 22
2 18
3 20
4 30
Explanation:
22 (2x)º + (3x - 5)º + (4x - 13)º = 180º [Angle sum property of triangle] 2x + 3x - 5 + 4x - 13 = 180º 9x - 18 = 180 9x = 198 x = 22 Hence, the correct answer is option (a).
THE TRIANGLE AND ITS PROPERTIES
298482
In Fig. if AB || CE, then the values of x and y are: 7
1 x = 26, y = 144
2 x = 36, y = 154
3 x = 154, y = 36
4 x = 144, y = 26
5 None of these.
Explanation:
None of these. \(\angle \text{DCA}+\angle \text{ACB}=180^\circ\) [Linear angles] \(\Rightarrow 134^\circ+\angle \text{ACB}=180^\circ\) \(\Rightarrow \angle \text{ACB}=46^\circ\) Now, In \(\triangle \text{ABC}\) \(\angle \text{BAC}+\angle \text{ACB}+\angle \text{ABC}=180^\circ\) [Angle sum property of triangle] \(\Rightarrow 62^\circ+46^\circ+2\text{x}^\circ=180^\circ\) \(\Rightarrow 2\text{x}^\circ=72^\circ\) \(\Rightarrow \text{x}=36\) Since, AB || CE \(\therefore \angle \text{ECB}=\angle \text{CBA}=(2\text{x})^\circ\) [Alternate angles] \(=(2\times 36)^\circ=72^\circ\) Now, \(\angle \text{DCE}+\angle \text{ECB}=180^\circ\) [Linear angles] \(\Rightarrow \text{y}^\circ+72^\circ=180^\circ\) \(\Rightarrow \text{y}=108\) Disclaimer: No Option is correct.
THE TRIANGLE AND ITS PROPERTIES
298483
Two angles of a triangle measure 90º and 30º. The measure of the third angle is.
1 90º
2 30º
3 60º
4 120º
Explanation:
60º Third angle = 180º - (90º + 30º) = 60º.
THE TRIANGLE AND ITS PROPERTIES
298484
Vikash wants to plant a flower on the ground in the form of a rhombus. The diagonals of the rhombus measure 42cm and 56cm. Find the perimeter of the field.
1 150cm
2 140cm
3 130cm
4 120cm
Explanation:
140cm Since diagonals of a rhombus bisect each other at 90° 8 Given, BD = 42cm and AC = 56cm Since BK \(=\frac{1}{2}\text{BD}=\frac{42}{2}=21\text{cm}\) Thus AK \(=\frac{1}{2}\text{AC}=\frac{56}{2}=28\text{cm}\) In \(\triangle\text{KAB}\) We have AB\(^{1}\) = AK\(^{1}\) + BK\(^{1}\) = (28)\(^{1}\) + (21)\(^{1}\) = 784 + 441 = 1225 Thus AB \(=\sqrt{1225}=35\text{cm}\) Therefore, perimeter of the field ABCD = 4 × 35 = 140cm
298481
If the measures of the angles of a triangle are (2x)º , (3x - 5)º and (4x -13)º. Then the value of x is:
1 22
2 18
3 20
4 30
Explanation:
22 (2x)º + (3x - 5)º + (4x - 13)º = 180º [Angle sum property of triangle] 2x + 3x - 5 + 4x - 13 = 180º 9x - 18 = 180 9x = 198 x = 22 Hence, the correct answer is option (a).
THE TRIANGLE AND ITS PROPERTIES
298482
In Fig. if AB || CE, then the values of x and y are: 7
1 x = 26, y = 144
2 x = 36, y = 154
3 x = 154, y = 36
4 x = 144, y = 26
5 None of these.
Explanation:
None of these. \(\angle \text{DCA}+\angle \text{ACB}=180^\circ\) [Linear angles] \(\Rightarrow 134^\circ+\angle \text{ACB}=180^\circ\) \(\Rightarrow \angle \text{ACB}=46^\circ\) Now, In \(\triangle \text{ABC}\) \(\angle \text{BAC}+\angle \text{ACB}+\angle \text{ABC}=180^\circ\) [Angle sum property of triangle] \(\Rightarrow 62^\circ+46^\circ+2\text{x}^\circ=180^\circ\) \(\Rightarrow 2\text{x}^\circ=72^\circ\) \(\Rightarrow \text{x}=36\) Since, AB || CE \(\therefore \angle \text{ECB}=\angle \text{CBA}=(2\text{x})^\circ\) [Alternate angles] \(=(2\times 36)^\circ=72^\circ\) Now, \(\angle \text{DCE}+\angle \text{ECB}=180^\circ\) [Linear angles] \(\Rightarrow \text{y}^\circ+72^\circ=180^\circ\) \(\Rightarrow \text{y}=108\) Disclaimer: No Option is correct.
THE TRIANGLE AND ITS PROPERTIES
298483
Two angles of a triangle measure 90º and 30º. The measure of the third angle is.
1 90º
2 30º
3 60º
4 120º
Explanation:
60º Third angle = 180º - (90º + 30º) = 60º.
THE TRIANGLE AND ITS PROPERTIES
298484
Vikash wants to plant a flower on the ground in the form of a rhombus. The diagonals of the rhombus measure 42cm and 56cm. Find the perimeter of the field.
1 150cm
2 140cm
3 130cm
4 120cm
Explanation:
140cm Since diagonals of a rhombus bisect each other at 90° 8 Given, BD = 42cm and AC = 56cm Since BK \(=\frac{1}{2}\text{BD}=\frac{42}{2}=21\text{cm}\) Thus AK \(=\frac{1}{2}\text{AC}=\frac{56}{2}=28\text{cm}\) In \(\triangle\text{KAB}\) We have AB\(^{1}\) = AK\(^{1}\) + BK\(^{1}\) = (28)\(^{1}\) + (21)\(^{1}\) = 784 + 441 = 1225 Thus AB \(=\sqrt{1225}=35\text{cm}\) Therefore, perimeter of the field ABCD = 4 × 35 = 140cm
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