In Fig. 10.11, find the value of x and y. -Maths 9th

Question

In Fig. 10.11, find the value of x and y. -Maths 9th

asked Feb 3, 2022 by anonymous

1 Answer

Answer :

Solution :- y = 2∠ ACB ⇒ y = 2 × 65° ⇒ y = 130° OA = OB (radii of the same circle) ∴ ∠ OBA = ∠ OAB ⇒ ∠ OBA = x In △ OAB, ∠ OAB + ∠ OBA + y = 180° x + x + 130° = 180° Or 2x = 50° Or x = 25°

answered Feb 3, 2022 by anonymous

Related questions

In the Fig.8.11. ABCD is a parallelogram.what is the sum of angles x,y and z? -Maths 9th

Description : In the Fig.8.11. ABCD is a parallelogram.what is the sum of angles x,y and z? -Maths 9th

Last Answer : Solution :-

For what value of x+y in fig.6.4 will ABC be a line? -Maths 9th

Description : For what value of x+y in fig.6.4 will ABC be a line? -Maths 9th

Last Answer : Solution :- For ABC to be a line, the sum of two adjacent angles must be 180°, i.e., x + y must be equal to 180°.

In Fig. 8.28, ABCD is a parallelogram. Find the value of x, y and z. -Maths 9th

Description : In Fig. 8.28, ABCD is a parallelogram. Find the value of x, y and z. -Maths 9th

Last Answer : Solution :-

In Fig. 6.10, if m|n, then find the value of x. -Maths 9th

Description : In Fig. 6.10, if m|n, then find the value of x. -Maths 9th

Last Answer : Solution :-

In Fig. 10.17, O is the centre of a circle, find the value of x. -Maths 9th

Description : In Fig. 10.17, O is the centre of a circle, find the value of x. -Maths 9th

Last Answer : Solution :- In △APB, ∠APB = 90° (Angle in the semi- circle) ∠APE + ∠ABP + ∠BAP = 180° 90° + 40° + ∠BAP = 180° ⇒ ∠BAP = 180° - 130° ⇒ ∠BAP = 50° Now, ∠BQP = ∠BAP (Angles in the same segment) ∴ x = 50°

From fig. 4.11 answer the following: -Maths 9th

Description : From fig. 4.11 answer the following: -Maths 9th

Last Answer : Solution :-

In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th

Description : In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th

Last Answer : Solution :-

In the Fig.8.7 ABCD is a | |gm.Find x and y. -Maths 9th

Description : In the Fig.8.7 ABCD is a | |gm.Find x and y. -Maths 9th

Last Answer : Solution :-

The sides BA and DC of quad. ABCD are produced as shown in Fig. 8.23. Prove that x + y = a + b -Maths 9th

Description : The sides BA and DC of quad. ABCD are produced as shown in Fig. 8.23. Prove that x + y = a + b -Maths 9th

Last Answer : Solution :-

In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Description : In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Last Answer : Solution :-

In Fig. 10.33, if OA = 10cm, AB = 16 cm and OD perpendicular to AB. Find the value of CD. -Maths 9th

Description : In Fig. 10.33, if OA = 10cm, AB = 16 cm and OD perpendicular to AB. Find the value of CD. -Maths 9th

Last Answer : Solution :- As OD is perpendicular to AB ⇒ AC = AB (Perpendicular from the centre to the chord bisects the chord) ∴ AC = AB/2 = 8cm In right △OCA, OA2 = AC2 + OC2 (102) = 82 + OC2 OC2 = 100 - 64 OC2 = 36 ... = 6cm CD = OD - OC = 10 - 6 = 4cm [∴ OA = OD = 10cm (radii)]

In Fig.6.6, find the value of x for which the lines l and m are parallel. -Maths 9th

Description : In Fig.6.6, find the value of x for which the lines l and m are parallel. -Maths 9th

Last Answer : Solution :-

In Fig. 6.7, if l||m, then find the value of x. -Maths 9th

Description : In Fig. 6.7, if l||m, then find the value of x. -Maths 9th

Last Answer : Solution :-

In Fig. 6.9, find the value of x. -Maths 9th

Description : In Fig. 6.9, find the value of x. -Maths 9th

Last Answer : Solution :-

In Fig. 6.17, AB| |CD. Find the value of x. -Maths 9th

Description : In Fig. 6.17, AB| |CD. Find the value of x. -Maths 9th

Last Answer : Solution :-

In Fig. 8.17, ABCD is a rhombus. Find the value of x. -Maths 9th

Description : In Fig. 8.17, ABCD is a rhombus. Find the value of x. -Maths 9th

Last Answer : Solution :-

In fig.4.7, LM is a line parallel to the y - axis at a distance of 2 units. -Maths 9th

Description : In fig.4.7, LM is a line parallel to the y - axis at a distance of 2 units. -Maths 9th

Last Answer : Given , LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units. (i) Coordinates of point P=(3,2) [since,its perpendicular distance from X-axis is 2] Coordinate of point ... L=3 , abscissa of point M=3 ∴∴ Difference between the abscissa of the points L and M =3-3=0

If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th

Description : If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th

Last Answer : Solution :- As, ∠ C = 90° (Angle in the semicircle) ∴ AC2 + BC2 = AB2 (By Pythagoras Theorem)

In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Description : In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Last Answer : Solution :- Jion AB ∠ABD = 90° (Angle in a semicircle) Similarly, ∠ABC = 90° So, ∠ABD + ∠ABC = 90° + 90° = 180° Therefore,DBC is a line i.e.,B lies on the segment DC.

In Fig. 10.25, a line intersect two concentric circles with centre O at A, B, C and D, Prove that AB = CD. -Maths 9th

Description : In Fig. 10.25, a line intersect two concentric circles with centre O at A, B, C and D, Prove that AB = CD. -Maths 9th

Last Answer : Solution :- Let OP be perpendicular from O on line l. Since the perpendicular from the centre of a circle to a chord,bisects the chord.Therefore, AP = DP ...(i) BP = CP ...(ii) Subtracting (ii) from (i), we get AP - BP = DP - CP ⇒ AB = CD

In Fig. 6.19, AB| |CD. Determine x. -Maths 9th

Description : In Fig. 6.19, AB| |CD. Determine x. -Maths 9th

Last Answer : Solution :-

In Fig. 6.20, find x if AB| |CD| |EF. -Maths 9th

Description : In Fig. 6.20, find x if AB| |CD| |EF. -Maths 9th

Last Answer : Solution :-

In Fig. 8.12, ABCD is a square. Find x. -Maths 9th

Description : In Fig. 8.12, ABCD is a square. Find x. -Maths 9th

Last Answer : Solution :-

For what value of p the point (p, 2) lies on the line 3x + y = 11? -Maths 9th

Description : For what value of p the point (p, 2) lies on the line 3x + y = 11? -Maths 9th

Last Answer : Solution :-

Write any four solutions of the linear equation y = 4 x – 11. -Maths 9th

Description : Write any four solutions of the linear equation y = 4 x – 11. -Maths 9th

Last Answer : answer:

What are the co-ordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 ? -Maths 9th

Description : What are the co-ordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 ? -Maths 9th

Last Answer : (d) 2x + 9y + 7 = 0PS being the median of ΔPQR, S is the mid-point of QR, i.e., Coordinates of S ≡ \(\bigg(rac{6+7}{2},rac{-1+3}{2}\bigg)\) = \(\bigg(rac{13}{2},1\bigg)\)Slope of line parallel to PS = Slope of PS= \(rac{1 ... y + 1) = \(rac{-2}{9}\)(x - 1), i.e., 9y + 9 = - 2x + 2 ⇒ 2x + 9y + 7 = 0.

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Last Answer : Given : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged ... length of coloured tape required = 30 cm (b) The values are : Happiness, beauty, Knowledge.

A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Description : A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Last Answer : Surface area of sphere = 4πr2, where r is the radius of sphere (ii) Height of cylinder, h = r+r =2r Radius of cylinder = r CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h) = 4πr2 (iii) Ratio ... sphere)/CSA of Cylinder) = 4r2/4r2 = 1/1 Ratio of the areas obtained in (i) and (ii) is 1:1.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Description : 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Last Answer : Q Solution: (i) In ΔAPB and ΔCQD, ∠ABP = ∠CDQ (Alternate interior angles) ∠APB = ∠CQD (= 90o as AP and CQ are perpendiculars) AB = CD (ABCD is a parallelogram) , ΔAPB ≅ ΔCQD [AAS congruency] (ii) As ΔAPB ≅ ΔCQD. , AP = CQ [CPCT]

Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus. -Maths 9th

Description : Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus. -Maths 9th

Last Answer : . Solution: (i) In ΔADC and ΔCBA, AD = CB (Opposite sides of a parallelogram) DC = BA (Opposite sides of a parallelogram) AC = CA (Common Side) , ΔADC ≅ ΔCBA [SSS congruency] Thus, ∠ACD = ∠CAB by ... are equal) Also, AB = BC = CD = DA (Opposite sides of a parallelogram) Thus, ABCD is a rhombus.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

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

Last Answer : Solution :-

From the fig. 4.8, write the following: -Maths 9th

Description : From the fig. 4.8, write the following: -Maths 9th

Last Answer : Solution:- (i) A =(3,2), B =(4,0), C =(0,4) (ii) D (iii) 2 (iv) -3

In Fig.5.6, if AC = BD, then prove that AB = CD. -Maths 9th

Description : In Fig.5.6, if AC = BD, then prove that AB = CD. -Maths 9th

Last Answer : if equals are subtracted to equals ,the remainders are equal subtract bc on both sides ab-bc=bd-bc ab=cd hence proved

In Fig.5.7, AC = XD, c is the mid-point of AB and D is the mid-point of XY. Using a Euclid's axiom,show that AB=XY. -Maths 9th

Description : In Fig.5.7, AC = XD, c is the mid-point of AB and D is the mid-point of XY. Using a Euclid's axiom,show that AB=XY. -Maths 9th

Last Answer : Solution :-

In Fig.6.5,which of the two lines are parallel? -Maths 9th

Description : In Fig.6.5,which of the two lines are parallel? -Maths 9th

Last Answer : Solution :- l||m, because angles on the same side of the transversal are supplementary, i.e., 128° +52° = 180°. Therefore p is not parallel to q, because 105° + 74° = 179°.

In Fig.6.29, DE||QR and AP and BP are bisectors of ZEAB and ZRBA respectively. Find ZAPB. -Maths 9th

Description : In Fig.6.29, DE||QR and AP and BP are bisectors of ZEAB and ZRBA respectively. Find ZAPB. -Maths 9th

Last Answer : Solution :-

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

ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Description : ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Last Answer : In △ABE and △ACF, we have ∠BEA=∠CFA (Each 90 0 ) ∠A=∠A (Common angle) AB=AC (Given) ∴△ABE≅△ACF (By SAS congruence criteria) ∴BF=CF [C.P.C.T]

In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th

Description : In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th

Last Answer : Solution :-

In Fig.8.6, BDEF and FDCE are parallelograms.Can you say that BD = CD? -Maths 9th

Description : In Fig.8.6, BDEF and FDCE are parallelograms.Can you say that BD = CD? -Maths 9th

Last Answer : Solution :-

Find all the angles of the | |gm ABCD given in the Fig. 8.8. -Maths 9th

Description : Find all the angles of the | |gm ABCD given in the Fig. 8.8. -Maths 9th

Last Answer : Solution :-

In Fig. 8.31, D is the mid-point of AB and PC = 1/2AP = 3 cm. If AD = DB = 4 cm and DE||BP. Find AE. -Maths 9th

Description : In Fig. 8.31, D is the mid-point of AB and PC = 1/2AP = 3 cm. If AD = DB = 4 cm and DE||BP. Find AE. -Maths 9th

Last Answer : Solution :-

In Fig. 8.32, ABCD and PQRB are rectangles where Q is the mid-point of BD. If QR = 5 cm, find the measure of AB. -Maths 9th

Description : In Fig. 8.32, ABCD and PQRB are rectangles where Q is the mid-point of BD. If QR = 5 cm, find the measure of AB. -Maths 9th

Last Answer : Solution :-

In Fig. 10.11, find the value of x and y. -Maths 9th

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

Answer :

Please log in or register to add a comment.

Related questions