Drawing Geometric Figures and Constructing Triangles: Grade 7 Math Worksheet Answer Key – PDF Download

Problem 1: Reading a protractor

The bottom side lies on the 0 of the outer scale, so read the outer scale where the other side crosses it. The side crosses at 65.

\(65^{\circ}\)

Problem 2: Triangle inequality check

Add the two shorter sides and compare the sum with the longest side.

1. \(3+4=7>5\): yes
2. \(6>2+3=5\): no
3. \(5+5=10\), which is not greater than 10: no
4. \(7+8=15>12\): yes

Problem 3: Third angle and type

(a) The angles of a triangle add to \(180^{\circ}\). \(48^{\circ} + 67^{\circ} = 115^{\circ}\), and \(180^{\circ} – 115^{\circ} = 65^{\circ}\).

(b) All three angles (\(48^{\circ}\), \(67^{\circ}\), \(65^{\circ}\)) are less than \(90^{\circ}\), so the triangle is acute.

Angle C = \(65^{\circ}\); the triangle is acute.

Problem 4: Angles of a triangle

The angles of a triangle must add up to \(180^{\circ}\).

1. \(50+60+70=180\): yes
2. \(90+60+40=190\): no
3. \(100+45+35=180\): yes
4. \(62+59+58=179\): no

Problem 5: Decimal side lengths

1. \(1.5 + 2.5 = 4\), which is not greater than 4. No.
2. \(2.5 + 4 = 6.5 > 6\). Yes.
3. \(3.2 + 3.2 = 6.4 > 6.3\). Yes.
4. \(0.8 + 1.1 = 1.9\), which is less than 2. No.

Problem 6: Third side length

The third side \(x\) must satisfy \(5+9>x\) and \(5+x>9\). So \(14>x\) and \(x>4\).

The whole numbers from 5 to 13 work. Smallest: 5 in. Largest: 13 in.

Problem 7: Isosceles bases

(a) The base must be greater than \(7 – 7 = 0\) and less than \(7 + 7 = 14\). Whole numbers: 1 to 13. The smallest is 1 in. and the largest is 13 in.

(b) The whole numbers from 1 through 13 give \(13 – 1 + 1 = 13\) bases. 13 bases.

Problem 8: Angles x, 2x, and 3x

The angles add to \(180^{\circ}\): \(x + 2x + 3x = 6x = 180\), so \(x = 180 : 6 = 30\).

The angles are \(30^{\circ}\), \(2 \times 30^{\circ} = 60^{\circ}\), and \(3 \times 30^{\circ} = 90^{\circ}\). Check: \(30 + 60 + 90 = 180\).

\(x = 30\); angles \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\); a right triangle.

Problem 9: Three angles

1. No. The bottom sides are 3 cm and 5 cm, so the second triangle is larger.
2. No. Three angles fix only the shape. Triangles of any size can have the same angles, so infinitely many triangles fit.

Problem 10: Equilateral triangle construction

(a) C is 5 cm from A and 5 cm from B, so \(AC = BC = 5\text{ cm}\).

(b) All three sides are equal, so the triangle is equilateral. Each angle measures \(60^{\circ}\), and \(3 \times 60^{\circ} = 180^{\circ}\).

(c) No. The arcs cross at one point above AB, so the classmate gets a triangle with the same size and shape. (The arcs also cross below AB, which gives a mirror image.)

AC = BC = 5 cm; each angle is \(60^{\circ}\); the triangle is unique.

Problem 11: Two sides and an angle between

Yes. The angle is between the two given sides. Once the 7 cm side and the \(40^{\circ}\) angle are drawn, point C is 5 cm along the second ray, and B and C are fixed. Only one triangle fits, so Sofia’s triangle matches Jamal’s.

Problem 12: Right angle between two sides

(a) A ruler measurement of BC gives 10 cm.

(b) Yes. Two sides and the angle between them are given, so exactly one triangle can be drawn.

(c) The sides are 6 cm, 8 cm, and 10 cm. All three are different, so the triangle is scalene.

Problem 13: One, many, or none

1. \(4 + 5 = 9 > 6\), so a triangle exists, and three sides fix its shape. Exactly one.
2. \(3 + 4 = 7\) is not greater than 8. No triangle.
3. The angles add to \(180^{\circ}\), so triangles exist, but no side length is given, so any size works. More than one.
4. The angles add to \(115^{\circ}\), less than \(180^{\circ}\), and the side AB is between them (ASA). Exactly one.
5. \(100 + 90 + 10 = 200\), which is more than \(180^{\circ}\). No triangle.

Problem 14: Two angles and a side

1. \(180-35-75=70\), so \(\angle C=70^{\circ}\).
2. The 8 cm side lies between the two angles, so the two rays meet at only one point C. Exactly one triangle.

Problem 15: Stick triangle

1. \(7+10=17\), which is less than 18. No, the two shorter sticks cannot reach across the third.
2. The new length \(x\) needs \(7+10=17>x\) and \(x>10-7=3\). The whole numbers from 4 to 16 work. Smallest: 4 in. Largest: 16 in.

Problem 16: Hiking loop

(a) CA must be greater than \(8 – 5 = 3\) and less than \(5 + 8 = 13\). The whole-number lengths are 4 through 12 miles.

(b) The loop is \(5 + 8 + CA = 13 + CA\). Shortest: \(13 + 4 = 17\) miles. Longest: \(13 + 12 = 25\) miles.

(c) \(20 – 13 = 7\), and 7 is between 4 and 12, so yes. CA = 7 miles.

Problem 17: Garden fence

1. The third side \(x\) satisfies \(40+25=65>x\) and \(x>40-25=15\). The whole numbers are 16 through 64 ft.
2. Smallest: \(40+25+16=81\). Largest: \(40+25+64=129\). Smallest 81 ft, largest 129 ft.

Problem 18: Land survey

(a) \(52^{\circ} + 71^{\circ} = 123^{\circ}\), and \(180^{\circ} – 123^{\circ} = 57^{\circ}\).

(b) Two angles and the side between them are given, so exactly one triangle fits.

(c) With \(60^{\circ}\) the angles would add to \(52 + 71 + 60 = 183\) degrees, which is more than \(180^{\circ}\). Angles of a triangle must add to exactly \(180^{\circ}\).

Angle at the tree: \(57^{\circ}\); one triangle; \(60^{\circ}\) is impossible.

Problem 19: Pennant banner

(a) With \(s = 6\), the two equal sides add to \(6 + 6 = 12\), which equals the base but is not greater than it. The sides would lie flat along the base, so no triangle forms.

(b) Need \(s + s > 12\), so \(s > 6\). The smallest whole number is 7 in.

(c) \(7 + 7 + 12 = 26\). The least perimeter is 26 in.

Problem 20: Two possible triangles

1. Two triangles: \(AB_1C\) and \(AB_2C\). Both have \(\angle A=30^{\circ}\), \(AC=10\text{ cm}\), and \(BC=7\text{ cm}\).
2. Triangle \(AB_2C\) has the longer side AB, about 13.6 cm, compared with about 3.8 cm for \(AB_1C\).
3. No. The given angle is not between the two given sides, so the arc can cross the ray twice, once, or not at all.

See every chapter on the Grade 7 math page.