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Post a LessonAnswered on 05 Mar Learn Practical Geometry
Sadika
To draw an equilateral triangle with one side length of 7 cm, follow these steps:
Draw a line segment of length 7 cm. This will represent one side of the equilateral triangle.
At one end of the line segment, mark point A.
With point A as the center and a radius of 7 cm, draw arcs on both sides of the line segment. These arcs should intersect the line segment at points B and C.
Connect points A, B, and C to form an equilateral triangle.
Now you have an equilateral triangle with one side length of 7 cm.
Answered on 05 Mar Learn Practical Geometry
Sadika
To draw triangle PQR with side lengths PQ = 3 cm, QR = 4 cm, and RP = 5 cm, as well as the bisector of angle Q, follow these steps:
Start by drawing a line segment QR of length 4 cm.
At one end of the line segment QR, mark point Q.
With Q as the center and a radius of 3 cm, draw an arc to intersect the line segment QR. Mark this point as P.
Now, with Q as the center and a radius of 5 cm, draw another arc to intersect the line segment QR. Mark this point as R.
Connect points P, Q, and R to form triangle PQR.
To draw the bisector of angle Q, first, find the midpoint of side QR. Let's call this point M.
Draw a straight line passing through point Q and the midpoint M of side QR. This line is the angle bisector of angle Q.
Now you have triangle PQR with side lengths PQ = 3 cm, QR = 4 cm, and RP = 5 cm, as well as the bisector of angle Q.
Answered on 05 Mar Learn Practical Geometry
Sadika
To draw a triangle with side lengths 4 cm, 5 cm, and 7 cm, and then draw the perpendicular bisector of the largest side, follow these steps:
Draw a line segment of length 7 cm. This will represent the longest side of the triangle.
At one end of the line segment, mark point A.
With A as the center and a radius of 5 cm, draw an arc to intersect the line segment. Mark this point as B.
Measure a distance of 4 cm from point B along the line segment BA. Mark this point as C.
Connect points A, B, and C to form triangle ABC.
Find the midpoint of side BC. Let's call this point M.
Draw a perpendicular line to side BC passing through the midpoint M of BC. This line is the perpendicular bisector of side BC.
Now you have a triangle ABC with side lengths 4 cm, 5 cm, and 7 cm, and the perpendicular bisector of the longest side BC.
Answered on 05 Mar Learn Practical Geometry
Sadika
To construct the bisector AD of angle A and the perpendicular AL from A on BC, and then measure angle LAD, follow these steps using only a ruler and compass:
Draw a line segment BC of length 7 cm.
At one end of BC, mark point B.
With B as the center and a radius of 6 cm, draw an arc to intersect the line segment BC. Mark this point as A.
With A as the center and a radius of 8 cm, draw an arc to intersect the line segment BC. Mark this point as C.
Connect points A, B, and C to form triangle ABC.
To construct the bisector of angle A (AD):
a. With compass, draw an arc from point A that intersects both sides of angle A.
b. Without changing the compass width, draw two arcs from the points where the first arc intersects the sides of angle A, creating two points of intersection on the arc.
c. Draw a straight line from point A through the point of intersection of the two arcs, extending beyond side BC. Label this line as AD.
To construct the perpendicular from A on BC (AL):
a. With compass width greater than half of BC (7 cm), draw two arcs centered at points B and C, intersecting each other above BC.
b. Label these intersection points as L and M.
c. Draw a straight line from point A through point L, perpendicular to BC. Label this line as AL.
Measure the angle LAD using a protractor.
Now, you have constructed the bisector AD of angle A and the perpendicular AL from A on BC. Measure the angle LAD using a protractor.
Answered on 05 Mar Learn Practical Geometry
Sadika
To draw triangle DEF with side lengths DE = DF = 4 cm and EF = 6 cm, follow these steps:
Draw a line segment of length 6 cm. This will represent side EF.
At one end of the line segment, mark point F.
With F as the center and a radius of 4 cm, draw an arc to intersect the line segment EF. Mark this point as D.
Now, with F as the center and a radius of 4 cm, draw another arc to intersect the line segment EF. Mark this point as E.
Connect points D, E, and F to form triangle DEF.
Now, to measure angles E and F, you can use a protractor:
Measure angle E: Place the center of the protractor at point E, align the baseline of the protractor along side DE, and read the angle where side EF intersects the protractor.
Measure angle F: Place the center of the protractor at point F, align the baseline of the protractor along side EF, and read the angle where side DE intersects the protractor.
Answered on 05 Mar Learn Practical Geometry
Sadika
Given: Length = 40 cm, Breadth = 22 cm
Perimeter of rectangle = 2*(length + breadth)
Perimeter of rectangle = 2*(40 cm + 22 cm) = 124 cm
Since the wire is bent to form a square, the perimeter of the square is equal to the perimeter of the rectangle:
Perimeter of square = 124 cm
Now, divide the perimeter of the square by 4 to get the side length of the square:
Side length of square = Perimeter of square / 4
Side length of square = 124 cm / 4 = 31 cm
Area of rectangle = length * breadth
Area of rectangle = 40 cm * 22 cm = 880 square cm
Area of square = (Side length of square)^2
Area of square = (31 cm)^2 = 961 square cm
Since 961 > 880, the square encloses more area.
read lessAnswered on 05 Mar Learn Practical Geometry
Sadika
Answered on 05 Mar Learn Practical Geometry
Sadika
Given: Wall dimensions = 3 m x 4 m, Tile dimensions = 10 cm x 12 cm
Area of wall = length * width
Area of wall = 3 m * 4 m = 12 square meters
Area of one tile = length * breadth
Area of one tile = 10 cm * 12 cm = 120 square cm
Number of tiles required = (Area of wall) / (Area of one tile)
Number of tiles required = 12 sq m * (10000 sq cm / 1 sq m) / 120 sq cm
Number of tiles required ≈ 100 tiles
Total cost of tiles = Number of tiles required * Rate per tile
Total cost of tiles = 100 tiles * Rs 2 = Rs 200
Answered on 05 Mar Learn Practical Geometry
Sadika
Given: Length = 9 dm 5 cm = 95 cm, Breadth = 6 dm 5 cm = 65 cm, Rate = 20 paise = Rs 0.20 per square cm
Area of table top = length * breadth
Area of table top = 95 cm * 65 cm = 6175 square cm
Cost of polishing = Area of table top * Rate
Cost of polishing = 6175 sq cm * Rs 0.20/sq cm = Rs 1235
Answered on 05 Mar Learn Practical Geometry
Sadika
Given: Room dimensions = 9.68 m x 6.2 m, Tile dimensions = 22 cm x 10 cm, Rate = Rs 2.50 per tile
Area of room floor = length * breadth
Area of room floor = 9.68 m * 6.2 m = 59.936 square meters
Area of one tile = length * breadth
Area of one tile = 22 cm * 10 cm = 220 square cm = 0.022 square meters
Number of tiles required = (Area of room floor) / (Area of one tile)
Number of tiles required = 59.936 sq m / 0.022 sq m = 2724.36 ≈ 2725 tiles
Total cost of tiles = Number of tiles required * Rate per tile
Total cost of tiles = 2725 tiles * Rs 2.50 = Rs 6812.50
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