Important Theorems on Triangles. -Maths 9th

Important Theorems on Triangles. -Maths 9th
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Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

Prove that a diagonal of a parallelogram divide it into two congruent triangles. -Maths 9th

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Last Answer : Given: A parallelogram ABCD and AC is its diagonal . To prove : △ABC ≅ △CDA Proof : In △ABC and △CDA, we have ∠DAC = ∠BCA [alt. int. angles, since AD | | BC] AC = AC [common side] and ∠BAC = ∠DAC [alt. int. angles, since AB | | DC] ∴ By ASA congruence axiom, we have △ABC ≅ △CDA

‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Description : On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Last Answer : According to question ∠BAC = ∠BDC.

How should i study maths main chapters like Lines and Angle,Triangles class 9 ?? -Maths 9th

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ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

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Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

Prove that a diagonal of a parallelogram divide it into two congruent triangles. -Maths 9th

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Last Answer : Given: A parallelogram ABCD and AC is its diagonal . To prove : △ABC ≅ △CDA Proof : In △ABC and △CDA, we have ∠DAC = ∠BCA [alt. int. angles, since AD | | BC] AC = AC [common side] and ∠BAC = ∠DAC [alt. int. angles, since AB | | DC] ∴ By ASA congruence axiom, we have △ABC ≅ △CDA

‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Description : On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Last Answer : According to question ∠BAC = ∠BDC.

In Fig. 4.6, if ABC and ABD are equilateral triangles then find the coordinates of C and D. -Maths 9th

Description : In Fig. 4.6, if ABC and ABD are equilateral triangles then find the coordinates of C and D. -Maths 9th

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If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Description : If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Last Answer : Solution :- △ ABC ≅ △PQR

If the corresponding angles of two triangles are equal, then they are always. State true or false and justify your answer. -Maths 9th

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Last Answer : Solution :- False, because two equilateral triangles with sides 3 cm and 6 cm respectively have all angles equal, but the triangles are not congruent.

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Last Answer : Linear equation in one variable: Linear equation of the form ax + b = 0, where a, b are real numbers such that a≠ 0. For example, 3x + 5 = 0. The letters used in an equation are called ... equation. More over every solution of the linear equation is a point on the graph of the linear equation.

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Last Answer : 1. In the figure, E is any point on median AD of a DABC. Show that ar (ABE) = ar (ACE). We have a ΔABC such that AD is a median. ∴ ar (ΔABD) = ar (ΔADC) ...(1) [∵ A median divides ... from both sides, we get [ar (ΔABD) - ar (ΔAOB)] = [ar (ΔABC) - ar (ΔAOB)] ⇒ ar (ΔAOD) = ar (ΔBOC)

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