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SSC (English Medium) इयत्ता १० वी - Maharashtra State Board Important Questions for Geometry Mathematics 2

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Geometry Mathematics 2
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∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that `"PQ"/"LT" = 3/4`.

Appears in 2 question papers
Chapter: [4] Geometric Constructions
Concept: Division of a Line Segment

Prove that “That ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.”

Appears in 2 question papers
Chapter: [4] Geometric Constructions
Concept: Geometric Constructions

Write down the equation of a line whose slope is 3/2 and which passes through point P, where P divides the line segment AB joining A(-2, 6) and B(3, -4) in the ratio 2 : 3.

Appears in 2 question papers
Chapter: [5] Co-ordinate Geometry
Concept: Division of a Line Segment

ΔRST ~ ΔUAY, In ΔRST, RS = 6 cm, ∠S = 50°, ST = 7.5 cm. The corresponding sides of ΔRST and ΔUAY are in the ratio 5 : 4. Construct ΔUAY.

Appears in 2 question papers
Chapter: [5] Co-ordinate Geometry
Concept: Division of a Line Segment

Construct the circumcircle and incircle of an equilateral ∆XYZ with side 6.5 cm and centre O. Find the ratio of the radii of incircle and circumcircle.

Appears in 2 question papers
Chapter: [5] Co-ordinate Geometry
Concept: Division of a Line Segment

∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that `"PQ"/"LT" = 3/4`.

Appears in 2 question papers
Chapter: [5] Co-ordinate Geometry
Concept: Division of a Line Segment

Find distance between point Q(3, – 7) and point R(3, 3)

Solution: Suppose Q(x1, y1) and point R(x2, y2)

x1 = 3, y1 = – 7 and x2 = 3, y2 = 3

Using distance formula,

d(Q, R) = `sqrt(square)`

∴ d(Q, R) = `sqrt(square - 100)`

∴ d(Q, R) =  `sqrt(square)`

∴ d(Q, R) = `square`

Appears in 2 question papers
Chapter: [5] Co-ordinate Geometry
Concept: Distance Formula

Prove that sin6θ + cos6θ = 1 – 3 sin2θ. cos2θ.

Appears in 2 question papers
Chapter: [6] Trigonometry
Concept: Trigonometric Identities (Square Relations)

Prove that:

sec2θ + cosec2θ = sec2θ x cosec2θ

Appears in 2 question papers
Chapter: [6] Trigonometry
Concept: Trigonometric Identities (Square Relations)

Prove that: If the angles of a triangle are 45° – 45° – 90°, then each of the perpendicular sides is \[\frac{1}{\sqrt{2}}\]times the hypotenuse.”

 

Appears in 2 question papers
Chapter: [6] Trigonometry
Concept: Angles in Standard Position

Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ

Appears in 2 question papers
Chapter: [6] Trigonometry
Concept: Trigonometric Identities (Square Relations)

Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`

Appears in 2 question papers
Chapter: [6] Trigonometry
Concept: Trigonometric Identities (Square Relations)

Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:

sin θ × cosec θ = ______

Appears in 2 question papers
Chapter: [6] Trigonometry
Concept: Trigonometric Identities (Square Relations)

Find the circumferences of a circle whose radius is 7 cm.

Appears in 2 question papers
Chapter: [7] Mensuration
Concept: Circumference of a Circle

The radii of the ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its curved surface area.

Appears in 2 question papers
Chapter: [7] Mensuration
Concept: Frustum of a Cone

The radius of a circle is 10 cm. The measure of an arc of the circle is 54°. Find the area of the sector associated with the arc. (\[\pi\]= 3.14 )

Appears in 2 question papers
Chapter: [7] Mensuration
Concept: Length of an Arc

In the given figure, square ABCD is inscribed in the sector A - PCQ. The radius of sector C - BXD is 20 cm. Complete the following activity to find the area of shaded region 

Appears in 2 question papers
Chapter: [7] Mensuration
Concept: Length of an Arc

The ratio of the areas of two triangles with the common base is 14 : 9. Height of the larger triangle is 7 cm, then find the corresponding height of the smaller triangle.

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Properties of Ratios of Areas of Two Triangles

The ratio of the areas of two triangles with common base is 6:5. Height of the larger triangle of 9 cm, then find the corresponding height of the smaller triangle.

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Properties of Ratios of Areas of Two Triangles

Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Properties of Ratios of Areas of Two Triangles
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