CBSE 10th Standard Maths Subject Introduction to Trigonometry Ncert Exemplar 2 Marks Questions With Solution 2021
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CBSE 10th Standard Maths Subject Introduction to Trigonometry Ncert Exemplar 2 Marks Questions With Solution 2021
10th Standard CBSE
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Reg.No. :
Maths
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Name the type of triangle formed by the points A(-5, 6), B(-4, 2) and C(7, 5).
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Find the value of m if the points (5, 1), (-2, -3) and (8, 2m) are collinear.
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The points A(2, 9), B(a, 5) and C(5, 5) Are the vertices of \(\triangle ABC\) right angled at B. Find the value of a and hence the area of \(\triangle ABC\).
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Name the type of triangle PQR formed by the points \(P(\sqrt { 2 } ,\sqrt { 2 } ),Q(-\sqrt { 2 } ,-\sqrt { 2 } )\) and \(R(-\sqrt { 6 } ,\sqrt { 6 } )\)
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Show that \(\Delta \)ABC with vertices A(-2, 0), B(0, 2) and C(2,0) is similar to \(\Delta \)DFE with vertices D(- 4, 0), E(4, 0) and F(0, 4).
2 Marks
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CBSE 10th Standard Maths Subject Introduction to Trigonometry Ncert Exemplar 2 Marks Questions With Solution 2021 Answer Keys
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scalene triangle
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\(m=\frac { 19 }{ 14 } \)
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\(a=2,\ OR\ \left( \triangle ABC \right) =6\ sq.units\)
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We have P(\(\sqrt{2}\) , \(\sqrt{2}\)) , Q(-\(\sqrt{2}\),-\(\sqrt{2}\)) and R(-\(\sqrt{6}\),\(\sqrt{6}\))
\(\therefore\) PQ = \(\sqrt { \left( \sqrt { 2 } +\sqrt { 2 } \right) ^{ 2 }+\left( \sqrt { 2 } +\sqrt { 2 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 2\sqrt { 2 } \right) ^{ 2 }+\left( 2\sqrt { 2 } \right) ^{ 2 } } \)
\(=\sqrt { 4\times 2+4\times 2 } =\sqrt { 8+8 } \)
\(=\sqrt { 16 } \)=4 units
PR = \(\sqrt { \left( \sqrt { 2 } +\sqrt { 6 } \right) ^{ 2 }+\left( \sqrt { 2 } +\sqrt { 6 } \right) ^{ 2 } } \)
\(=\sqrt { 2+6+2\sqrt { 2 } +2+6-2\sqrt { 2 } } \)
\(=\sqrt { 2+6+2+6 } \)
\(=\sqrt { 16 } \) = 4 units
RQ = \(\sqrt { [(-\sqrt { 2 } )+\sqrt { 6 } ^{ 2 }+\left( -\sqrt { 2 } -\sqrt { 6 } \right) ^{ 2 } } \)
\(=\sqrt { 2+6-2\sqrt { 2 } +2+6+2\sqrt { 2 } } \)
\(=\sqrt { 2+6+2+6 } \)
\(=\sqrt { 16 } \) =4 units
Since PQ=PR=RQ = 4 units
∴ PQR is an equilateral triangle. -
Given, vertices of \(\Delta \)ABC are A(-2, 0), B(0, 2) and C(2,0) and D(- 4, 0), E(4, 0) and F(0, 4).
Now, AB = \(\sqrt { { (0+2) }^{ 2 }+{ (2-0) }^{ 2 } } =\sqrt { 4+4 } =2\sqrt { 2 } \) units [\(\because \) distance=\(\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \)]
BC = \(\sqrt { { (2-0) }^{ 2 }+{ (0-2) }^{ 2 } } \) \( =\sqrt { 4+4 } =2\sqrt { 2 } \) units
CA = \(\sqrt { { (-2-2) }^{ 2 }+{ (0-0) }^{ 2 } } \) = \(\sqrt { { (-4) }^{ 2 }+0 } \)= 4 units
FD = \(\sqrt { { (0+4) }^{ 2 }+{ (4-0) }^{ 2 } } \) = \(\sqrt { { (4) }^{ 2 }+{ (-4) }^{ 2 } } \) = \(4\sqrt { 2 } \) units
FE = \(\sqrt { { (4-0) }^{ 2 }+{ (0-4) }^{ 2 } } \) = \(\sqrt { { (4) }^{ 2 }+{ (-4) }^{ 2 } } \) = \(4\sqrt { 2 } \) units
and ED = \(\sqrt { { (-4-4) }^{ 2 }+{ (0-0) }^{ 2 } } \) = \(\sqrt { { (-8) }^{ 2 }} \) = \(\sqrt { {64} }\) = 8 units
Here, we see that sides of \(\Delta \) DEF are twice the sides of \(\Delta \)ABC.
Hence, both the triangle are similar.
Hence proved.
2 Marks
